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  • php edit database entery script not working.

    - by Jacksta
    I have two files which show all contats in a database, then edits them. The second script keeps directing the browser back due to this part. I cant work out how to make it work! :) <?php if (!$_POST[id]) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION[valid] != "yes") { header( "Location: pick_modcontact.php"); exit; } This script shows all contacts in a database which is an "address book" this part works fine. please see below. Name:pick_modcontact.php if ($_SESSION[valid] != "yes") { header( "Location: contact_menu.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "user") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = @mysql_num_rows($result); if ($num < 1) { $display_block = "<p><em>Sorry No Results!</em></p>"; } else { while ($row = mysql_fetch_array($result)) { $id = $row['id']; $f_name = $row['f_name']; $l_name = $row['l_name']; $option_block .= "<option value\"$id\">$f_name, $l_name</option>"; } $display_block = "<form method=\"POST\" action=\"show_modcontact.php\"> <p><strong>Contact:</strong> <select name=\"id\">$option_block</select> <input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p> </form>"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <h1>My Contact Management System</h1> <h2><em>Modify a Contact</em></h2> <p>Select a contact from the list below, to modify the contact's record.</p> <? echo "$display_block"; ?> <br> <p><a href="contact_menu.php">Return to Main Menu</a></p> </body> </html> This script is for modifying the contact: named show_modcontact.php <?php if (!$_POST[id]) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION[valid] != "yes") { header( "Location: pick_modcontact.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin_cantsayno", "cantsayno") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT f_name, l_name, address1, address2, address3, postcode, prim_tel, sec_tel, email, birthday FROM $table_name WHERE id = '$_POST[id]'"; $result = @mysql_query($sql, $connection) or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $f_name = $row['f_name']; $l_name = $row['l_name']; $address1 = $row['address1']; $address2 = $row['address2']; $address3 = $row['address3']; $country = $row['country']; $prim_tel = $row['prim_tel']; $sec_tel = $row['sec_tel']; $email = $row['email']; $birthday = $row['birthday']; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <form action="do_modcontact.php" method="post"> <input type="id" value="<? echo "$_POST[id]"; ?>" /> <table cellpadding="5" cellspacing="3"> <tr> <th>Name & Address Information</th> <th> Other Contact / Personal Information</th> </tr> <tr> <td align="top"> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name" ?>" size="35" maxlength="75" /></p> <p><strong>Last Name:</strong><br /> <input type="text" name="l_name" value="<? echo "$l_name" ?>" size="35" maxlength="75" /></p> <p><strong>Address1:</strong><br /> <input type="text" name="f_name" value="<? echo "$address1" ?>" size="35" maxlength="75" /></p> <p><strong>Address2:</strong><br /> <input type="text" name="f_name" value="<? echo "$address2" ?>" size="35" maxlength="75" /></p> <p><strong>Address3:</strong><br /> <input type="text" name="f_name" value="<? echo "$address3" ?>" size="35" maxlength="75" /> </p> <p><strong>Postcode:</strong><br /> <input type="text" name="f_name" value="<? echo "$postcode" ?>" size="35" maxlength="75" /></p> <p><strong>Country:</strong><br /> <input type="text" name="f_name" value="<? echo "$country" ?>" size="35" maxlength="75" /> </p> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name" ?>" size="35" maxlength="75" /></p> </td> <td align="top"> <p><strong>Prim Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$prim_tel" ?>" size="35" maxlength="75" /></p> <p><strong>Sec Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$sec_tel" ?>" size="35" maxlength="75" /></p> <p><strong>Email:</strong><br /> <input type="text" name="f_name" value="<? echo "$email" ?>" size="35" maxlength="75" /> </p> <p><strong>Birthday:</strong><br /> <input type="text" name="f_name" value="<? echo "$birthday" ?>" size="35" maxlength="75" /> </p> </td> </tr> <tr> <td align="center"> <p><input type="submit" name="submit" value="Update Contact" /></p> <br /> <p><a href="contact_menu.php">Retuen To Menu</a></p> </td> </tr> </table> </form> </body> </html>

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  • php cookie not working

    - by Jacksta
    I am trying to use a cookie with authentication. This page works once entering user and pass <? if ((!$_POST[username]) || (!$_POST[password])) { header("Location: show_login.html"); exit; } $db_name = "testDB"; $table_name = "auth_users"; $connection = @mysql_connect("localhost", "user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT * FROM $table_name WHERE username ='$_POST[username]' AND password = password('$_POST[password]')"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = mysql_num_rows($result); if ($num != 0) { $cookie_name = "auth"; $cookie_value = "ok"; $cookie_expire = "0"; $cookie_domain = "domain.com.au"; setcookie($cookie_name, $cookis_value, $cookie_expire, "/", $cookie_domain, 0); $display_block = " <p><strong>Secret Menu:</strong></p> <ul> <li><a href=\"secretA.php\">secret page A</a> <li><a href=\"secretB.php\">secret page B</a> </ul>"; } else { header("Location: show_login.html"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Secret Area</title> </head> <body> <? echo "$display_block"; ?> </body> </html> WHen clicking on either secretA.php or secretB.php I am redirected to log in again, it should work. here is the code. secretB.php <?php if ($_COOKIE[auth] == "ok") { $msg = "<p>Welcome to secret page B, authorised user! </P>"; } else { header( "Location: /show_login.html"); exit; } ?> <HTML> <HEAD> <title>Secret Page B:</title> </HEAD> <BODY> <? echo "$msg"; ?> </BODY> </HTML>

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  • create new db in mysql with php syntax

    - by Jacksta
    I am trying to create a new db called testDB2, below is my code. Once running the script all I am getting an error on line 7 Fatal error: Call to undefined function mysqlquery() in /home/admin/domains/domain.com.au/public_html/db_createdb.php on line 7 This is my code <? $sql = "CREATE database testDB2"; $connection = mysql_connect("localhost", "admin_user", "pass") or die(mysql_error()); $result = mysqlquery($sql, $connection) or die(mysql_error()); if ($result) { $msg = "<p>Databse has been created!</p>"; } ?> <HTML> <head> <title>Create MySQL database</title> </head> <body> <? echo "$msg"; ?> </body> </HTML>

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  • php file path of server

    - by Jacksta
    I am trying to get this script to work. it opens up a directry and lists the files in the directory. I have copied this code from somewhere else and the problem is that this php file is hosted on an apache server not my localhost. what is the correct $dir_name = "c:/"; to use? The file is in this directory /domains/domainxxxx.com.au/public_html/lsitfiles.php so would I use domainxxxx.com.au/public_html/lsitfiles.php ? <?php $dir_name = "c:/"; $dir = opendir($dir_name); $file_list = "<ul>"; while ($file_name = readdir($dir)) { if(($file_name != ".") && (file_name != "..")) { $file_list .= "<li>$file_name"; } } $file_list.= "<ul>"; closedir($dir); ?> <HTML> <BODY> <p>Files in: <? echo "$dir_name"; ?></p> <? echo "$file_list"; ?> </BODY> </HTML>

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  • what is the point of escaping quotation marks in php

    - by Jacksta
    Here is a validation script from a book I am learning, Why is escaping the quotation marks necassery? e.g. <option value=\"char\">char</option> <?php //validate important input if ((!$_POST[table_name]) || (!$_POST[num_fields])) { header( "location: show_createtable.html"); exit; } //begin creating form for display $form_block = " <form action=\"do_createtable.php\" method=\"post\"> <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\"> <table cellspacing=\"5\" cellpadding=\"5\"> <tr> <th>Field Name</th><th>Field Type</th><th>Table Length</th> </tr>"; //count from 0 until you reach the number fo fields for ($i = 0; $i <$_POST[num_fields]; $i++) { $form_block .=" <tr> <td align=center><input type=\"texr\" name=\"field name[]\" size=\"30\"></td> <td align=center> <select name=\"field_type[]\"> <option value=\"char\">char</option> <option value=\"date\">date</option> <option value=\"float\">float</option> <option value=\"int\">int</option> <option value=\"text\">text</option> <option value=\"varchar\">varchar</option> </select> </td> <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"> </td> </tr>"; } //finish up the form $form_block .= " <tr> <td align=center colspan=3><input type =\"submit\" value=\"create table\"> </td> </tr> </table> </form>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create a database table: Step 2</title> </head> <body> <h1>defnie fields for <? echo "$_POST[table_name]"; ?> </h1> <? echo "$form_block"; ?> </body> </html>

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  • php upload file function

    - by Jacksta
    I am trying to write a script which uploads a file via a html form. When I click submit nothing happens. file: upload_form.html <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <form action="do_upload.php" method="post" enctype="multipart/form-data"></form> <p><strong>File to upload</strong></p> <p><input name="img1" type="file" size="30" /></p> <p><input name="submit" type="submit" value="Upolad File" /></p> </body> </html> file: do_upload.php <?php if ($_FILES[img1] != "" { @copy($_FILES[img1] [tm_name], "/tmp" .$_FILES[img1][name]) or die("couldnt copy the file"); } else { die("no file specified"); } ?> <HTML> <head> <title>Successfull File Upload</title> </head> <body> <h1>Success</h1> <p>You sent: <? echo $_FILES[img1][name]; ?>, a <? echo $_FILES[img1][size]; ?>byte filw with a mime type of <? echo $_FILES[img1][type]; ?></p> </body> </HTML>

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  • php MySQL snytax error

    - by Jacksta
    my scrip is supposed to look up contacts in a table and present thm on the screen to then be edited. however this is not this case. I am getting the error Parse error: syntax error, unexpected $end in /home/admin/domains/domain.com.au/public_html/pick_modcontact.php on line 50 NOTE: this is the last line in this script. <? session_start(); if ($_SESSION[valid] != "yes") { header( "Location: contact_menu.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = @mysql_num_rows($result); if ($num < 1) { $display_block = "<p><em>Sorry No Results!</em></p>"; } else { while ($row = mysql_fetch_array($result)) { $id = $row['id']; $f_name = $row['f_name']; $l_name = $row['l_name']; $option_block .= "<option value\"$id\">$l_name, $f_name</option>"; } $display_block = "<form method=\"POST\" action=\"show_modcontact.php\"> <p><strong>Contact:</strong> <select name=\"id\">$option_block</select> <input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p> </form>"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <h1>My Contact Management System</h1> <h2><em>Modify a Contact</em></h2> <p>Select a contact from the list below, to modify the contact's record.</p> <? echo "$display_block"; ?> <br> <p><a href="contact_menu.php">Return to Main Menu</a></p> </body> </html>

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  • setting a cookie in php

    - by Jacksta
    I am trying to set a cookie, whas wrong with this as I am getting an error. Warning: setcookie() expects parameter 3 to be long, string given in /home/admin/domains/domain.com.au/public_html/setcookie.php on line 6 <?php $cookie_name = "test_cookie"; $cookie_value = "test_string"; $cookie_expire = "time()+86400"; $cookie_domain = "localhost"; setcookie($cookie_name, $cookis_value, $cookie_expire, "/", $cookie_domain, 0); ?> <HTM> <HEAD> </HEAD> <BODY> <h1>cookie mmmmmmm</h1> </BODY> </HTML>

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  • php and mysql listing databases and looping through results

    - by Jacksta
    Beginner help needed :) I am doign an example form a php book which lists tables in databases. I am getting an error on line 36: $db_list .= "$table_list"; <?php //connect to database $connection = mysql_connect("localhost", "admin_cantsayno", "cantsayno") or die(mysql_error()); //list databases $dbs = @mysql_list_dbs($connection) or die(mysql_error()); //start first bullet list $db_list = "<ul>"; $db_num = 0; //loop through results of functions while ($db_num < mysql_num_rows($dbs)) { //get database names and make each a list point $db_names[$db_num] = mysql_tablename($dbs, $db_num); $db_list .= "<li>$db_names[$db_num]"; //get table names and make another list $tables = @mysql_list_tables($db_names[$db_num]) or die(mysql_error()); $table_list = "<ul>"; $table_num = 0; //loop through results of function while ($table_num < mysql_num_rows($tables)){ //get table names and make each bullet point $table_names[$table_num] = mysql_tablename($tables, $table_num); $table_list .= "<li>$table_names[$table_num]"; $table_num++; } //close inner bullet list and increment number to continue $table_list .= "</ul>" $db_list .= "$table_list"; $db_num++; } //close outer bullet list $db_list .= "</ul>"; ?> <html> <head> <title>MySQL Tables</title> </head> <body> <p><strong>Data bases and tables on local host</strong></p> <? echo "$db_list"; ?> </body>

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  • facebook application testing

    - by Jacksta
    I am goign to start mucking around with a facebook app. You are spposed to be able to test code here http://developers.facebook.com/tools.php but that url is a 404. Does anyone know if the Test Console still exists?

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  • php and mysql site design question

    - by Jacksta
    I am trying to build a website with mysql and php. This is the first site I have attempted so I want to write a little plan and get some feedback. The site allows users to add some text in a text field as a “comment”. Once the comment has been entered into the site it is added to the database where it can be voted for by other users. When a new comment has been added to the database it needs to create a new page, e.g. www.xxxxx.com/commentname or www.xxxxxx.com/?id=99981. There will be a list of "Comments" in the database along with the number of votes for each comment. The home page will have two functions. 1) Allow users to add a "comment" 2) Display two tables, each with 20 rows containing most "popular comments" and "recent comments" Each comment will generate its one page where the comment will be displayed. Here users can read the comment and Vote for the comment if they wish. Please help me out by explaining how to do the following. -Generate a new page whenever a comment is added to the database -Add a vote to the vote count in the comment database. -Display the top 20 most popular comments as per number of votes.

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  • mysql_connect()

    - by Jacksta
    I am trying to connect to mysql and am getting an error. I put my servers ip address in and used port 3306 whihch post should be used? <?php $connection = mysql_connect("serer.ip:port", "user", "pass") or die(mysql_error()); if ($connection) {$msg = "success";} ?> <html> <head> </head> <body> <? echo "$msg"; ?> </body> </html> Here is the error its producing Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'admin'@'server1.myserver.com' (using password: YES) in /home/admin/domains/mydomain.com.au/public_html/db_connect.php on line 3 Access denied for user 'admin'@'server1.myserver.com' (using password: YES)

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  • php syntax error

    - by Jacksta
    I have 3 files 1) show_createtable.html 2) do_showfielddef.php 3) do_showtble.php 1) First file is for creating a new table for a data base, it is a fom with 2 inputs, Table Name and Number of Fields. THIS WORKS FINE! <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <h1>Step 1: Name and Number</h1> <form method="post" action="do_showfielddef.php" /> <p><strong>Table Name:</strong><br /> <input type="text" name="table_name" size="30" /></p> <p><strong>Number of fields:</strong><br /> <input type="text" name="num_fields" size="30" /></p> <p><input type="submit" name="submit" value="go to step2" /></p> </form> </body> </html> 2) this script validates fields and createa another form to enter all the table rows. This for also WORKS FINE! <?php //validate important input if ((!$_POST[table_name]) || (!$_POST[num_fields])) { header( "location: show_createtable.html"); exit; } //begin creating form for display $form_block = " <form action=\"do_createtable.php\" method=\"post\"> <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\"> <table cellspacing=\"5\" cellpadding=\"5\"> <tr> <th>Field Name</th><th>Field Type</th><th>Table Length</th> </tr>"; //count from 0 until you reach the number fo fields for ($i = 0; $i <$_POST[num_fields]; $i++) { $form_block .=" <tr> <td align=center><input type=\"texr\" name=\"field name[]\" size=\"30\"></td> <td align=center> <select name=\"field_type[]\"> <option value=\"char\">char</option> <option value=\"date\">date</option> <option value=\"float\">float</option> <option value=\"int\">int</option> <option value=\"text\">text</option> <option value=\"varchar\">varchar</option> </select> </td> <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"> </td> </tr>"; } //finish up the form $form_block .= " <tr> <td align=center colspan=3><input type =\"submit\" value=\"create table\"> </td> </tr> </table> </form>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create a database table: Step 2</title> </head> <body> <h1>defnie fields for <? echo "$_POST[table_name]"; ?> </h1> <? echo "$form_block"; ?> </body> </html> Problem is here 3) this form creates the tables and enteres them into the database. I am getting an error on line 37 "Parse error: syntax error, unexpected $end in /home/admin/domains/domaina.com.au/public_html/do_createtable.php on line 37" <? $db_name = "testDB"; $connection = @mysql_connect("localhost", "admin_user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "CREATE TABLE $_POST[table_name]("; for ($i = 0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[field_length][$i] !="") { $sql .=" (".$_POST[field_length][$i]."),"; } else { $sql .=","; } $sql = substr($sql, 0, -1); $sql .= ")"; $result = mysql_query($sql, $connection) or die(mysql_error()); if ($result) { $msg = "<p>" .$_POST[table_name]." has been created!</p>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create A Database Table: Step 3</title> </head> <body> <h1>Adding table to <? echo "$db_name"; ?>...</h1> <? echo "$msg"; ?> </body> </html>

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  • PHP syntax error “unexpected $end”

    - by Jacksta
    I have 3 files 1) show_createtable.html 2) do_showfielddef.php 3) do_showtble.php 1) First file is for creating a new table for a data base, it is a fom with 2 inputs, Table Name and Number of Fields. THIS WORKS FINE! <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <h1>Step 1: Name and Number</h1> <form method="post" action="do_showfielddef.php" /> <p><strong>Table Name:</strong><br /> <input type="text" name="table_name" size="30" /></p> <p><strong>Number of fields:</strong><br /> <input type="text" name="num_fields" size="30" /></p> <p><input type="submit" name="submit" value="go to step2" /></p> </form> </body> </html> 2) this script validates fields and createa another form to enter all the table rows. This for also WORKS FINE! <?php //validate important input if ((!$_POST[table_name]) || (!$_POST[num_fields])) { header( "location: show_createtable.html"); exit; } //begin creating form for display $form_block = " <form action=\"do_createtable.php\" method=\"post\"> <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\"> <table cellspacing=\"5\" cellpadding=\"5\"> <tr> <th>Field Name</th><th>Field Type</th><th>Table Length</th> </tr>"; //count from 0 until you reach the number fo fields for ($i = 0; $i <$_POST[num_fields]; $i++) { $form_block .=" <tr> <td align=center><input type=\"texr\" name=\"field name[]\" size=\"30\"></td> <td align=center> <select name=\"field_type[]\"> <option value=\"char\">char</option> <option value=\"date\">date</option> <option value=\"float\">float</option> <option value=\"int\">int</option> <option value=\"text\">text</option> <option value=\"varchar\">varchar</option> </select> </td> <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"> </td> </tr>"; } //finish up the form $form_block .= " <tr> <td align=center colspan=3><input type =\"submit\" value=\"create table\"> </td> </tr> </table> </form>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create a database table: Step 2</title> </head> <body> <h1>defnie fields for <? echo "$_POST[table_name]"; ?> </h1> <? echo "$form_block"; ?> </body> </html> Problem is here 3) this form creates the tables and enteres them into the database. I am getting an error on line 37 "Parse error: syntax error, unexpected $end in /home/admin/domains/domaina.com.au/public_html/do_createtable.php on line 37" <? $db_name = "testDB"; $connection = @mysql_connect("localhost", "admin_user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "CREATE TABLE $_POST[table_name]("; for ($i = 0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[field_length][$i] !="") { $sql .=" (".$_POST[field_length][$i]."),"; } else { $sql .=","; } $sql = substr($sql, 0, -1); $sql .= ")"; $result = mysql_query($sql, $connection) or die(mysql_error()); if ($result) { $msg = "<p>" .$_POST[table_name]." has been created!</p>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create A Database Table: Step 3</title> </head> <body> <h1>Adding table to <? echo "$db_name"; ?>...</h1> <? echo "$msg"; ?> </body> </html>

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  • authorise user from mysql database

    - by Jacksta
    I suck at php, and cant find the error here. The script gets 2 variables "username" and "password" from a html from then check them against a MySQL databse. When I run this I get the follow error "Query was empty" <? if ((!$_POST[username]) || (!$_POST[password])) { header("Location: show_login.html"); exit; } $db_name = "testDB"; $table_name = "auth_users"; $connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $slq = "SELECT * FROM $table_name WHERE username ='$_POST[username]' AND password = password('$_POST[password]')"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = mysql_num_rows($result); if ($num != 0) { $msg = "<p>Congratulations, you're authorised!</p>"; } else { header("Location: show_login.html"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Secret Area</title> </head> <body> <? echo "$msg"; ?> </body> </html>

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  • php + MySQL editing table data.

    - by Jacksta
    This question is relating to 2 php scripts. The first script is called pick_modcontact.php where I choose a contact (from a contact book like phone book), then posts to the script show_modcontact.php When I click the submit button on the form on pick.modcontact.php. As a result of submitting the form I am then taken to show_modcontact.php. As the variables are not present the user is directed back to pick_modcontact.php I can not work out how to correct the code so that it will show the results of the script show_modcontact.php This script shows all contacts in a database which is an "address book" this part works fine. please see below. Name:pick_modcontact.php if ($_SESSION['valid'] != "yes") { header( "Location: contact_menu.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "user") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = @mysql_num_rows($result); if ($num < 1) { $display_block = "<p><em>Sorry No Results!</em></p>"; } else { while ($row = mysql_fetch_array($result)) { $id = $row['id']; $f_name = $row['f_name']; $l_name = $row['l_name']; $option_block .= "<option value\"$id\">$f_name, $l_name</option>"; } $display_block = "<form method=\"POST\" action=\"show_modcontact.php\"> <p><strong>Contact:</strong> <select name=\"id\">$option_block</select> <input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p> </form>"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <h1>My Contact Management System</h1> <h2><em>Modify a Contact</em></h2> <p>Select a contact from the list below, to modify the contact's record.</p> <? echo "$display_block"; ?> <br> <p><a href="contact_menu.php">Return to Main Menu</a></p> </body> </html> This script is for modifying the contact: named show_modcontact.php <?php if (!$_POST['id']) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION['valid'] != "yes") { header( "Location: pick_modcontact.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT f_name, l_name, address1, address2, address3, postcode, prim_tel, sec_tel, email, birthday FROM $table_name WHERE id = '" . $_POST['id'] . "'"; $result = @mysql_query($sql, $connection) or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $f_name = $row['f_name']; $l_name = $row['l_name']; $address1 = $row['address1']; $address2 = $row['address2']; $address3 = $row['address3']; $country = $row['country']; $prim_tel = $row['prim_tel']; $sec_tel = $row['sec_tel']; $email = $row['email']; $birthday = $row['birthday']; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <form action="do_modcontact.php" method="post"> <input type="text" name="id" value="<? echo $_POST['id']; ?>" /> <table cellpadding="5" cellspacing="3"> <tr> <th>Name & Address Information</th> <th> Other Contact / Personal Information</th> </tr> <tr> <td align="top"> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name"; ?>" size="35" maxlength="75" /></p> <p><strong>Last Name:</strong><br /> <input type="text" name="l_name" value="<? echo "$l_name"; ?>" size="35" maxlength="75" /></p> <p><strong>Address1:</strong><br /> <input type="text" name="f_name" value="<? echo "$address1"; ?>" size="35" maxlength="75" /></p> <p><strong>Address2:</strong><br /> <input type="text" name="f_name" value="<? echo "$address2"; ?>" size="35" maxlength="75" /></p> <p><strong>Address3:</strong><br /> <input type="text" name="f_name" value="<? echo "$address3"; ?>" size="35" maxlength="75" /> </p> <p><strong>Postcode:</strong><br /> <input type="text" name="f_name" value="<? echo "$postcode"; ?>" size="35" maxlength="75" /></p> <p><strong>Country:</strong><br /> <input type="text" name="f_name" value="<? echo "$country"; ?>" size="35" maxlength="75" /> </p> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name"; ?>" size="35" maxlength="75" /></p> </td> <td align="top"> <p><strong>Prim Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$prim_tel"; ?>" size="35" maxlength="75" /></p> <p><strong>Sec Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$sec_tel"; ?>" size="35" maxlength="75" /></p> <p><strong>Email:</strong><br /> <input type="text" name="f_name" value="<? echo "$email;" ?>" size="35" maxlength="75" /> </p> <p><strong>Birthday:</strong><br /> <input type="text" name="f_name" value="<? echo "$birthday"; ?>" size="35" maxlength="75" /> </p> </td> </tr> <tr> <td align="center"> <p><input type="submit" name="submit" value="Update Contact" /></p> <br /> <p><a href="contact_menu.php">Retuen To Menu</a></p> </td> </tr> </table> </form> </body> </html> note for site admin, I am re posting this question with the hope of someone else reading over it. older questions seem to go dead after a while.

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  • Script to create a table and fields in SQL wont work

    - by Jacksta
    Warning this is lenghty! attack if you knowledagble. well at least more then a newb beginner like me. This script uses three files as detailed below. It is suppoed to create the database and fields from the form input. It gets to the end and shows my_contacts has been created!. But when i go into phpMyadmin the table has not been created. I have a file named show_createtable.html which is used to create a table in MySQL <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <h1>Step 1: Name and Number</h1> <form method="post" action="do_showfielddef.php" /> <p><strong>Table Name:</strong><br /> <input type="text" name="table_name" size="30" /></p> <p><strong>Number of fields:</strong><br /> <input type="text" name="num_fields" size="30" /></p> <p><input type="submit" name="submit" value="go to step2" /></p> </form> </body> </html> This Form Posts to do_showfielddef.php <?php //validate important input if ((!$_POST[table_name]) || (!$_POST[num_fields])) { header( "location: show_createtable.html"); exit; } //begin creating form for display $form_block = " <form action=\"do_createtable.php\" method=\"post\"> <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\"> <table cellspacing=\"5\" cellpadding=\"5\"> <tr> <th>Field Name</th><th>Field Type</th><th>Table Length</th><th>Primary Key?</th><th>Auto-Increment?</th> </tr>"; //count from 0 until you reach the number fo fields for ($i = 0; $i <$_POST[num_fields]; $i++) { $form_block .=" <tr> <td align=center><input type=\"texr\" name=\"field name[]\" size=\"30\"></td> <td align=center> <select name=\"field_type[]\"> <option value=\"char\">char</option> <option value=\"date\">date</option> <option value=\"float\">float</option> <option value=\"int\">int</option> <option value=\"text\">text</option> <option value=\"varchar\">varchar</option> </select> </td> <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"></td> <td aligh=center><input type=\"checkbox\" name=\"primary[]\" value=\"Y\"></td> <td aligh=center><input type=\"checkbox\" name=\"auto_increment[]\" value=\"Y\"></td> </tr>"; } //finish up the form $form_block .= " <tr> <td align=center colspan=3><input type =\"submit\" value=\"create table\"> </td> </tr> </table> </form>"; ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create a database table: Step 2</title> </head> <body> <h1>defnie fields for <? echo "$_POST[table_name]"; ?> </h1> <? echo "$form_block"; ?> </body> </html> Which in turn creates the table and fields with this file do_showfielddef.php //connect to database $connection = @mysql_connect("localhost", "user", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); //start creating the SQL statement $sql = "CREATE TABLE $_POST[table_name]("; //continue the SQL statement for each new field for ($i = 0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[auto_increment][$i] =="Y") { $additional = "NOT NULL auto_increment"; } else { $additional = ""; } if ($_POST[primary][$i] =="Y") { $additional .= ", primary key (".$_POST[field_name][$i].")"; } else { $additional = ""; } if ($_POST[field_length][$i] !="") { $sql .= " (".$_POST[field_length][$i].") $additional ,"; } else { $sql .=" $additional ,"; } } //clean up the end of the string $sql = substr($sql, 0, -1); $sql .= ")"; //execute the query $result = mysql_query($sql, $connection) or die(mysql_error()); //get a giid message for display upon success if ($result) { $msg = "<p>" .$_POST[table_name]." has been created!</p>"; } ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Create A Database Table: Step 3</title> </head> <body> <h1>Adding table to <? echo "$db_name"; ?>...</h1> <? echo "$msg"; ?> </body> </html>

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  • Unexpected T_ELSE error in PHP

    - by Jacksta
    I am working on an example from a php book and am getting an error on line 8 with this code <?php $agent = getenv("HTTP_USER_AGENT"); if (preg_match("/MSIE/i", "$agent")); { $result = "You are using Microsoft Internet Explorer"; } else if (preg_match("/Mozilla/i", "$agent")); { $result = "You are using Mozilla firefox"; } else {$result = "you are using $agent"; } echo $result; ?>

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  • How to use the php fopen() correctly

    - by Jacksta
    I am learning php, trying to use the fopen() function. The php file I am coding is in this directory /domains/xxxxx.com.au/public_html/phpfile.php What path do I specify for the file to be opened, the example I am looking at is based on a server on a pc where this is the file path $filename = "c:/newfile.txt"; not an online server.

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