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  • Connecting form to database errors

    - by Russell Ehrnsberger
    Hello I am trying to connect a page to a MySQL database for newsletter signup. I have the database with 3 fields, id, name, email. The database is named newsletter and the table is named newsletter. Everything seems to be fine but I am getting this error Notice: Undefined index: Name in C:\wamp\www\insert.php on line 12 Notice: Undefined index: Name in C:\wamp\www\insert.php on line 13 Here is my form code. <form action="insert.php" method="post"> <input type="text" value="Name" name="Name" id="Name" class="txtfield" onblur="javascript:if(this.value==''){this.value=this.defaultValue;}" onfocus="javascript:if(this.value==this.defaultValue){this.value='';}" /> <input type="text" value="Enter Email Address" name="Email" id="Email" class="txtfield" onblur="javascript:if(this.value==''){this.value=this.defaultValue;}" onfocus="javascript:if(this.value==this.defaultValue){this.value='';}" /> <input type="submit" value="" class="button" /> </form> Here is my insert.php file. <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="newsletter"; // Database name $tbl_name="newsletter"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // Get values from form $name=$_POST['Name']; $email=$_POST['Email']; // Insert data into mysql $sql="INSERT INTO $tbl_name(name, email)VALUES('$name', '$email')"; $result=mysql_query($sql); // if successfully insert data into database, displays message "Successful". if($result){ echo "Successful"; echo "<BR>"; echo "<a href='index.html'>Back to main page</a>"; } else { echo "ERROR"; } ?> <?php // close connection mysql_close(); ?>

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  • opening and viewing a file in php

    - by Christian Burgos
    how do i open/view for editing an uploaded file in php? i have tried this but it doesn't open the file. $my_file = 'file.txt'; $handle = fopen($my_file, 'r'); $data = fread($handle,filesize($my_file)); i've also tried this but it wont work. $my_file = 'file.txt'; $handle = fopen($my_file, 'w') or die('Cannot open file: '.$my_file); $data = 'This is the data'; fwrite($handle, $data); what i have in mind is like when you want to view an uploaded resume,documents or any other ms office files like .docx,.xls,.pptx and be able to edit them, save and close the said file. edit: latest tried code... <?php // Connects to your Database include "configdb.php"; //Retrieves data from MySQL $data = mysql_query("SELECT * FROM employees") or die(mysql_error()); //Puts it into an array while($info = mysql_fetch_array( $data )) { //Outputs the image and other data //Echo "<img src=localhost/uploadfile/images".$info['photo'] ."> <br>"; Echo "<b>Name:</b> ".$info['name'] . "<br> "; Echo "<b>Email:</b> ".$info['email'] . " <br>"; Echo "<b>Phone:</b> ".$info['phone'] . " <hr>"; //$file=fopen("uploadfile/images/".$info['photo'],"r+"); $file=fopen("Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt","r") or exit("unable to open file");; } ?> i am getting the error: Warning: fopen(Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt): failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/uploadfile/view.php on line 17 unable to open file the file is in that folder, i don't know it wont find it.

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  • Not allowing characters after Space. Mysql Insert With PHP

    - by Jake
    Ok so I think this is easy but I dont know (I'm a novice to PHP and MySQL). I have a select that is getting data from a table in the database. I am simply taking whatever options the user selects and putting it into a separate table with a php mysql insert statement. But I am having a problem. When I hit submit, everything is submitted properly except for any select options that have spaces don't submit after the first space. For example if the option was COMPUTER REPAIR, all that would get sent is COMPUTER. I will post code if needed, and any help would be greatly appreciated. Thanks! Ok here is my select code: <?php include("./config.php"); $query="SELECT id,name FROM category_names ORDER BY name"; $result = mysql_query ($query); echo"<div style='overflow:auto;width:100%'><label>Categories (Pick three that describe your business)</label><br/><select name='select1'><option value='0'>Please Select A Category</option>"; // printing the list box select command while($catinfo=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option>$catinfo[name]</option><br/> "; } echo"</select></div>"; ?> And here is my insert code ( Just to let you know its got everything not just the select!) ?php require("./config.php"); $companyname = mysql_real_escape_string(addslashes(trim($_REQUEST['name']))); $phone = mysql_real_escape_string(addslashes($_REQUEST['phone'])); $zipcode = mysql_real_escape_string(addslashes($_REQUEST['zipcode'])); $city = mysql_real_escape_string(addslashes($_REQUEST['city'])); $description = mysql_real_escape_string(addslashes($_REQUEST['description'])); $website = mysql_real_escape_string(addslashes($_REQUEST['website'])); $address = mysql_real_escape_string(addslashes($_REQUEST['address'])); $other = mysql_real_escape_string(addslashes($_REQUEST['other'])); $payment = mysql_real_escape_string(addslashes($_REQUEST['payment'])); $products = mysql_real_escape_string(addslashes($_REQUEST['products'])); $email = mysql_real_escape_string(addslashes($_REQUEST['email'])); $select1 = mysql_real_escape_string(addslashes($_REQUEST['select1'])); $select2 = mysql_real_escape_string(addslashes($_REQUEST['select2'])); $select3 = mysql_real_escape_string(addslashes($_REQUEST['select3'])); $save=$_POST['save']; if(!empty($save)){ $sql="INSERT INTO gj (name, phone, city, zipcode, description, dateadded, website, address1, other2, payment_options, Products, email,cat1,cat2,cat3) VALUES ('$companyname','$phone','$city','$zipcode','$description',curdate(),'$website','$address','$other','$payment','$products','$email','$select1','$select2','$select3')"; if (!mysql_query($sql,$link)) { die('Error: ' . mysql_error()); } echo "<br/><h2><font color='green' style='font-size:15px'>1 business added</font></h2>"; mysql_close($link); } ?>

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  • How to print arguments passed to configure script?

    - by Sam
    Hi, I'm trying to print arguments passed to a ./configure script. Calling 'echo' on $BASH_ARGV will just print the last set of arguments. For example if I run: ./configure --enable-foo --enable-bar echo $BASH_ARGV will print only "--enable-bar" How do I print all the arguments? Thanks!

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  • Ajax call not working...

    - by Probocop
    Hi, I have a form that submits to a PHP script with Jquery and Ajax. The PHP script returns some XML. For some reason the Ajax success function is not firing, and the error ones is. Can anybody see where I'm going wrong? My Jquery is as follows $('#submit-excuse').submit(function (event) { event.preventDefault(); ws_url = 'http://jacamo.epiphanydev2.co.uk/content/inc/excuse-submit.php?excuse='+$('input#excuse').val(); $.ajax({ type: 'GET', url: ws_url, dataType: "xml", beforeSend: function() { $('p#response').text('Sending.'); }, success: function(xmlIn) { results = xmlIn.getElementsByTagName("ReportID"); }, error: function() { $('p#response').text('Error.'); } }); }); And my PHP script is as follows: $excuse = $_GET['excuse']; $badwords = array ( 'one', 'two', 'three', 'four', 'five' ); if ($excuse == '') { $error = 'enter something'; } else { foreach ($badwords as $word) { $pos = strpos($excuse, $word); if($pos !== false) { $passed = false; } } if ($passed !== false) { $username = 'xxxxx'; $password = 'xxxxx'; $message = $excuse; $url = 'http://twitter.com/statuses/update.xml'; $curl_handle = curl_init(); curl_setopt($curl_handle, CURLOPT_URL, "$url"); curl_setopt($curl_handle, CURLOPT_CONNECTTIMEOUT, 2); curl_setopt($curl_handle, CURLOPT_RETURNTRANSFER, 1); curl_setopt($curl_handle, CURLOPT_POST, 1); curl_setopt($curl_handle, CURLOPT_POSTFIELDS, "status=$message"); curl_setopt($curl_handle, CURLOPT_USERPWD, "$username:$password"); $buffer = curl_exec($curl_handle); curl_close($curl_handle); $passed = 'yes'; } echo "<?xml version='1.0' encoding='UTF-8'?>\n"; echo "\t<result>\n"; echo "\t\t<passed>" . $passed . "</passed>\n"; echo "\t</result>"; } Thanks

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  • What's the best way/practice to get the extension of a uploaded file in PHP

    - by Roland
    I have a form that allow users to upload files, I however need to get the file extension, which I am able to get, but not sure if I'm using the most effective solution I can get it using the following ways $fileInfo = pathinfo($_FILES['File']['name']); echo $fileInfo['extension']; $ext = end(explode('.',$_FILES['File']['name'])); echo $ext; Which method is the best to use or are there even better solutions that would get the extension?

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  • how to increment a javascript variable title that is within a php while loop

    - by steve
    I'm building multiple countdown clocks on one page. The number of countdown clocks varies from day to day so I need to call javascript several times from within "while" code in php to produce different clocks. The following code works but it's based on knowing how many clocks are needed before I start: <script language="javascript" src="countdown.js"></script> <script language="javascript"> var cd1 = new countdown('cd1'); cd1.Div = "clock1"; cd1.TargetDate = "<?php echo "$clocktime"; ?>"; cd1.DisplayFormat = "%%D%% days, %%H%% hours, %%M%% minutes, %%S%% seconds until event AAA happens"; </script> <div id="clockwrapper"><div id="clock1">[clock]</div></div> <script language="javascript" src="countdown.js"></script> <script language="javascript"> var cd2 = new countdown('cd2'); cd2.Div = "clock2"; cd2.TargetDate = "02/01/2011 5:30:30 PM"; cd2.DisplayFormat = "%%D%% days, %%H%% hours, %%M%% minutes, %%S%% seconds until event BBB happens..."; </script> <div id="clockwrapper"><div id="clock2">[clock]</div></div> So if I keep on calling the javascript above (the code with cd1 in it) all previous "cd1" clocks change to the latest clock because it is being overwritten. Somehow I need to call javascript from within my "while" loop in php and have cd1 become cd2, then cd3 so that the clocks work as they're supposed to. How do I go about doing this? I don't know how to call the javascript several times and increment the variable cd1 within the javascript. I tried something like this but couldn't get it to work. $id=mysql_result($result,$i,"id"); while($id){ $cd = ("$cd"."$id"); ?> <script language="javascript" src="countdown.js"></script> <script language="javascript"> var <?php echo "$cd"; ?> = new countdown('<?php echo "$cd"; ?>'); .... </script> <div id="clockwrapper"><div id="<?php echo "$cd"; ?>">[clock]</div></div> <?php $id=mysql_result($result,$i,"id"); } ?> Surely there is some easy way of getting around this that I don't know about. Thanks

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  • Trouble echoing data - data being converted to 0

    - by Haraldo
    Hi, I feel this is something to do with my httpd setup for apache. I'm using mod_rewrite if that helps but I think that only effects the url. It seems when I output some data such as: $sMessage = 'Error'; echo $sMessage; It works fine but when I do this: $sMessage = 'Error'; echo ''+$sMessage+''; It returns 0. Very odd!

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  • [bash] checking wget's return value [if]

    - by wwrob
    I'm writing a script to download a bunch of files, and I want it to inform when a particular file doesn't exist. r=`wget -q www.someurl.com` if [ $r -ne 0 ] then echo "Not there" else echo "OK" fi But it gives the following error on execution: ./file: line 2: [: -ne: unary operator expected What's wrong?

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  • shell script array length

    - by Dipro Sen
    I assume arguments to my shell scripts willbe ./x.sh subject N file1 file2 fileN So I am splicing argv from 3 till end candidates=${@:3} now I want to check whether length of candidates is same as given N I am trying with echo $((${#candidates[@]})) which is always returning 1. I can do echo "$#-2" | bc but, I shouldn't I be able to get array size ? I can use bc to do integer comparison. but I've to know the size of `candidates array which I am not getting properly.

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  • I need to copy only selected files and folders in PHP

    - by OM The Eternity
    I am using the following code, in which initially i am taking the difference of two folder structure and then the out put needs to be copied to other folder. here is the code below.. $source = '/var/www/html/copy1'; $mirror = '/var/www/html/copy2'; function scan_dir_recursive($dir, $rel = null) { $all_paths = array(); $new_paths = scandir($dir); foreach ($new_paths as $path) { if ($path == '.' || $path == '..') { continue; } if ($rel === null) { $path_with_rel = $path; } else { $path_with_rel = $rel . DIRECTORY_SEPARATOR . $path; } $full_path = $dir . DIRECTORY_SEPARATOR . $path; $all_paths[] = $path_with_rel; if (is_dir($full_path)) { $all_paths = array_merge( $all_paths, scan_dir_recursive($full_path, $path_with_rel) ); } } return $all_paths; } $diff_paths = array_diff( scan_dir_recursive($mirror), scan_dir_recursive($source) ); /*$diff_path = array_diff($mirror,$original);*/ echo "<pre>Difference ";print_r($diff_paths); foreach($diff_paths as $path) { echo $source1 = "var/www/html/copy2/".$path; echo "<br>"; $des = "var/www/html/copy1/".$path; copy_recursive_dirs($source1, $des); } function copy_recursive_dirs($dirsource, $dirdest) { $dir_handle=opendir($dirsource); mkdir($dirdest,0777); while(false!==($file=readdir($dir_handle))) {/*echo "<pre>"; print_r($file);*/ if($file!="." && $file!="..") { if(is_dir($dirsource.DIRECTORY_SEPARATOR.$file)) { //Copy the file at the same level in the destination folder copy_recursive_dirs($dirsource.DIRECTORY_SEPARATOR.$file, $dirdest.DIRECTORY_SEPARATOR.$file); } else{ //Copy the dir at the same lavel in the destination folder copy ($dirsource.DIRECTORY_SEPARATOR.$file, $dirdest.DIRECTORY_SEPARATOR.$file); } } } closedir($dir_handle); return true; } Whenever I execute the script I get the difference output but do not get the other copy on second folder as per code... Pls help me in rectifying...

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  • What does Apache need to support both mysqli and PDO?

    - by Nathan Long
    I'm considering changing some PHP code to use PDO for database access instead of mysqli (because the PDO syntax makes more sense to me and is database-agnostic). To do that, I'd need both methods to work while I'm making the changeover. My problem is this: so far, either one or the other method will crash Apache. Right now I'm using XAMPP in Windows XP, and PHP Version 5.2.8. Mysqli works fine, and so does this: $dbc = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password); echo 'Connected to database'; $sql = "SELECT * FROM `employee`"; But this line makes Apache crash: $dbc->query($sql); I don't want to redo my entire Apache or XAMPP installation, but I'd like for PDO to work. So I tried updating libmysql.dll from here, as oddvibes recommended here. That made my simple PDO query work, but then mysqli queries crashed Apache. (I also tried the suggestion after that one, to update php_pdo_mysql.dll and php_pdo.dll, to no effect.) Test Case I created this test script to compare PDO vs mysqli. With the old copy of libmysql.dll, it crashes if $use_pdo is true and doesn't if it's false. With the new copy of libmysql.dll, it's the opposite. if ($use_pdo){ $dbc = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password); echo 'Connected to database<br />'; $sql = "SELECT * FROM `employee`"; $dbc->query($sql); foreach ($dbc->query($sql) as $row){ echo $row['firstname'] . ' ' . $row['lastname'] . "<br>\n"; } } else { $dbc = @mysqli_connect($hostname, $username, $password, $dbname) OR die('Could not connect to MySQL: ' . mysqli_connect_error()); $sql = "SELECT * FROM `employee`"; $result = @mysqli_query($dbc, $sql) or die(mysqli_error($dbc)); while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) { echo $row['firstname'] . ' ' . $row['lastname'] . "<br>\n"; } } What does Apache need in order to support both methods of database query?

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  • Using /dev/tcp instead of wget

    - by User1
    Why does this work: exec 3</dev/tcp/www.google.com/80 echo -e "GET / HTTP/1.1\n\n"&3 cat <&3 And this fail: echo -e "GET / HTTP/1.1\n\n" /dev/tcp/www.google.com/80 cat </dev/tcp/www.google.com/80 Is there a way to do it in one-line w/o using wget, curl, or some other library?

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  • Linking Post Title to Specific Page ID

    - by ThatMacLad
    I've created a form to update my websites homepage with content but I wanted to know how I could set it up so that a posts title links to a specific post ID. I'd also like to add a Read More link that directs anybody reading the blog to the correct post. Here is my PHP code: <html> <head> <title>Blog Name</title> </head> <body> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "SELECT * FROM php_blog ORDER BY timestamp DESC LIMIT 5"; $result = mysql_query($sql) or print ("Can't select entries from table php_blog.<br />" . $sql . "<br />" . mysql_error()); while($row = mysql_fetch_array($result)) { $date = date("l F d Y", $row['timestamp']); $title = stripslashes($row['title']); $entry = stripslashes($row['entry']); $password = $row['password']; $id = $row['id']; if ($password == 1) { echo "<p><strong>" . $title . "</strong></p>"; printf("<p>This is a password protected entry. If you have a password, log in below.</p>"); printf("<form method=\"post\" action=\"post.php?id=%s\"><p><strong><label for=\"username\">Username:</label></strong><br /><input type=\"text\" name=\"username\" id=\"username\" /></p><p><strong><label for=\"pass\">Password:</label></strong><br /><input type=\"password\" name=\"pass\" id=\"pass\" /></p><p><input type=\"submit\" name=\"submit\" id=\"submit\" value=\"submit\" /></p></form>",$id); print "<hr />"; } else { ?> <p><strong><?php echo $title; ?></strong><br /><br /> <?php echo $entry; ?><br /><br /> Posted on <?php echo $date; ?> <hr /></p> <?php } } ?> </body> </html> Thanks for any help. I really appreciate any input!

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  • Classes within classes in PHP

    - by Matt
    Can you do this in PHP? I've heard conflicting opinions: Something like: Class bar { function a_function () { echo "hi!"; } } Class foo { public $bar; function __construct() { $this->bar = new bar(); } } $x = new foo(); $x->bar->a_function(); Will this echo "hi!" or not?

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  • alternative of `die()` in php

    - by Prasoon Saurav
    I have the following script <?php echo "I am alive<br>"; die("I am dying<br>"); echo ("Dead"); ?> The output that I get is I am alive I am dying Is there any way (alternative/substitute of die()) using which the execution of the remaining script be continued?

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  • How to get base url with php?

    - by shin
    I am using XAMPP on windows vista. In my development, I have http://127.0.0.1/test_website/ Now I would like get this http://127.0.0.1/test_website/ with php. I tried something like these, but none of them worked. echo dirname(__FILE__) or echo basename(__FILE__); etc. I will appreciate any help. Thanks in advance.

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  • how can I count number of occurance of a word in a string while ignoring cases

    - by Ronny
    I am trying to count the number of occurance of a given word while ignoring cases. I have tried <?php $string = 'Hello World! EARTh in earth and EARth';//string to look into. if(stristr($string, 'eartH')) { echo 'Found';// this just show eartH was found. } $timesfound = substr_count($string, stristr($string, 'eartH'));// this count how many times. echo $timesfound; // this outputs 1 instead of 3.

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  • Float conditional in bash

    - by Werner
    Hi, in bash I need to compare two float numbers, one which I define in the script and the other read as paramter, for that I do: if [[ $aff -gt 0 ]] then a=b echo "xxx "$aff #echo $CX $CY $CZ $aff fi but I get the error: [[: -309.585300: syntax error: invalid arithmetic operator (error token is ".585300") What is wrong? Thanks

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