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  • What's wrong with my If-statement to check uploaded file? (PHP)

    - by ggfan
    I am trying to test if the uploaded file is the image type I want. If it isn't a gif,jpeg, png, it should echo "Problem". But when I execute this code, it always says there's a problem. What's wrong with my if statement? $uploadfile_type=$_FILES['userfile']['type']; if ( ($uploadfile_type !='image/gif') || ($uploadfile_type !='image/jpeg') || ($uploadfile_type !='image/png')) { echo 'Problem: file is not a gif or jpeg or png!'; exit; } This code works when I am only checking one type of image. Ex: if($uploadfile_type !='image/gif') -- this statement would work but when I add a OR it doesn't.

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  • Stopping users posting more than once

    - by user342391
    Before posting my form I am checking the database to see if there are any previous posts from the user. If there are previous posts then the script will kick back a message saying you have already posted. The problem is that what I am trying to achieve isn't working it all goes wrong after my else statement. It is also probable that there is an sql injection vulnerability too. Can you help??4 <?php include '../login/dbc.php'; page_protect(); $customerid = $_SESSION['user_id']; $checkid = "SELECT customerid FROM content WHERE customerid = $customerid"; if ($checkid = $customerid) {echo 'You cannot post any more entries, you have already created one';} else $sql="INSERT INTO content (customerid, weburl, title, description) VALUES ('$_POST[customerid]','$_POST[webaddress]','$_POST[pagetitle]','$_POST[pagedescription]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?>

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  • Prolem with if function

    - by Ryan
    Hi, something seems to be wrong with the first line of this if function, seems alright to me though. if ($count1 == sizeof($map) && $count2 == sizeof($map[0])){ echo ";"; }else{ echo ","; } This is the error I get (line 36 is the first line of the above line.) Parse error: parse error in C:\wamp\www\game\mapArrayConvertor.php on line 36 EDIT: The OP notes in an answer below that the error was a missing semi-colon on line 35 and not the code included in the question.

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  • returning values from function or method multiple times by only calling the class once

    - by Sokhrat Sikar
    I have a members.php file that shows my websites members. I echo members name by using foreach method. A method of Members class returns an array, then I use foreach loop in the members.php file to echo the members. I am trying to aovid writing php code in my members.php file. Is there a way to avoid using foreach inside members.php file? For example, is it possible to return value from a method couple of times? (by only calling the object once). Just like how we normally call the functions? This question doesn't make sense, but I am just trying to see if there is a away around this issue?

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  • show count of rows between 2 dates

    - by hello
    I am trying to show the number of rows that have a created_at date between 2 dates. Here is my code: $result=mysql_query("select * from payments where created_at between '2013/10/01 00:00:00' and '2013/10/30 00:00:00'") or die('You need to add an administrator ' ); $counter = mysql_query("select * from payments where created_at between '2013/10/01 00:00:00' and '2013/10/30 00:00:00'"); $row = mysql_fetch_array($result); $id = $row['id']; $num = mysql_fetch_array($counter); $countjan = $num["id"]; However when i echo (<?php echo"$jan";?>)this shows as 0 any idea how i can get this to work P.s there is 1 row within this date range

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  • CodeIgniter - Returning multiple file upload details

    - by Chris
    Hey All, Im using the codeigniter upload library to upload multiple files, which works fine ... What im having problems with is returning the information about the files. Im using the following code to print the results for testing echo '<pre>'; print_r($this->upload->data()); echo '</pre>'; A cut down version of the results are as follows Array ( [file_name] => Array ( [0] => filename1.gif [1] => filename2.jpg ) ) The way my view is setup, is that i use jquery to insert multiple dynamic file input fields so the amount of files can be 1, it can be 50 and so on. Im wondering how i would loop through that array to send each filename to the database

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  • How place a link below a picture that is displayed using fancybox (jquery)?

    - by janoChen
    In the first picture of my website (the two persons), shows an URL address when you click on it. I used the "title" thing. Is there a simple way of doing the same but placing a link instead? code: <div class="pusher"> <h3><?php echo l('showcase1_h3'); ?></h3> <p><?php echo l('showcase1_p'); ?></p> <div class="pic"> <a id="showcase1" title="studyatbest.com" href="images/showcase1.png"><img src="images/showcase1t.png"/></a> </div> </div> http://alexchen.co.nr/

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  • php multidimensional array if loop

    - by user1091558
    I have a multidimensional array like this array[value][1][1] Now i would like to implement if loop like this if ($value = array[value][1][1]) { echo "It works"; } Now it works if i assign the values like [1][1],[2][1]. Is it possible to compare the whole array. I mean if the array looks like array[value][1][1],array[value][2][1],..........,array[value][n][1] It works should be echoed. I tried like this. if ($value = array[value][][]) { echo "It works"; } But its not working. Can anyone give me the correct syntax?

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  • Proper mechanism for sending PHP errors to the client

    - by Chris
    Greetings, I was trying to discover a proper way to send captured errors or business logic exceptions to the client in an Ajax-PHP system. In my case, the browser needs to react differently depending on whether a request was successful or not. However in all the examples I've found, only a simple string is reported back to the browser in both cases. Eg: if (something worked) echo "Success!"; else echo "ERROR: that failed"; So when the browser gets back the Ajax response, the only way to know if an error occurred would be to parse the string (looking for 'error' perhaps). This seems clunky. Is there a better/proper way to send back the Ajax response & notify the browser of an error? Thank you.

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  • convert mysql code to codeigniter

    - by Jethro Tamares Doble
    How can i convert this code into an acceptable codeigniter code: mysql_select_db($database_connection_ched, $connection_ched); $query_Institutions = "SELECT * FROM tb_institutional_profile ORDER BY tb_institutional_profile.institution_name ASC"; $Institutions = mysql_query($query_Institutions, $connection_ched) or die(mysql_error()); $row_Institutions = mysql_fetch_assoc($Institutions); $totalRows_Institutions = mysql_num_rows($Institutions); <td width="192"><select name="institution_id"> <?php do { <option value="<?php echo $row_Institutions['institution_id']?>" ><?php echo $row_Institutions['institution_name']?></option> <?php } while ($row_Institutions = mysql_fetch_assoc($Institutions)); ?> </select></td>

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  • list of all commits SHA1_HASH

    - by dorelal
    git init echo 'I am foo' > foo.txt git add foo.txt # this woould create a git commit object echo ' I am foo2' > foo.txt git add foo.txt # this would create another git commit object git commit -m 'doe' # this would create two git # objects: one commit object and one tree object How do I get a list of all 4 commits SHA1_HASH ? cd .git/objects ls (master)$ ls -al total 0 drwxr-xr-x 8 nsingh staff 272 Mar 27 16:44 . drwxr-xr-x 13 nsingh staff 442 Mar 27 16:44 .. drwxr-xr-x 3 nsingh staff 102 Mar 27 16:44 37 drwxr-xr-x 3 nsingh staff 102 Mar 27 16:43 a2 drwxr-xr-x 3 nsingh staff 102 Mar 27 16:44 e1 drwxr-xr-x 3 nsingh staff 102 Mar 27 16:42 e6 drwxr-xr-x 2 nsingh staff 68 Mar 27 16:42 info drwxr-xr-x 2 nsingh staff 68 Mar 27 16:42 pack I can find the list of all 4 commits by looking at file here but there must be a better way.

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  • Is it possible for a parent to use a child's constant or static variables from inside a static metho

    - by ryeguy
    Below is an example of what I'm trying to do. The parent can't access any of the child's variables. It doesn't matter what technique I use (static or constants), I just need some kind of functionality like this. class ParentClass { public static function staticFunc() { //both of these will throw a (static|const) not defined error echo self::$myStatic; echo self::MY_CONSTANT; } } class ChildClass extends ParentClass { const MY_CONSTANT = 1; public static $myStatic = 2; } ChildClass::staticFunc();

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  • php/mySQL error: mysql_num_rows(): supplied argument is not a valid MySQL result

    - by Michael Robinson
    I'm trying to INSERT INTO a mySQL database and I'm getting this error on: if (mysql_num_rows($login) == 1){ Here is the php, The php does add the user to the database. I can't figure it out. <? session_start(); require("config.php"); $u = $_GET['username']; $pw = $_GET['password']; $pwh = $_GET['passwordhint']; $em = $_GET['email']; $zc = $_GET['zipcode']; $check = "INSERT INTO family (loginName, email, password, passwordhint, location) VALUES ('$u', '$pw', '$pwh', '$em', '$zc')"; $login = mysql_query($check, $link) or die(mysql_error()); if (mysql_num_rows($login) == 1) { $row = mysql_fetch_assoc($login); echo 'Yes';exit; } else { echo 'No';exit; } mysql_close($link); ?> Thanks,

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  • Can't use method return value in write context; Not sure where to go from here

    - by Morgan Green
    This is my source for the variable. <?php if ($admin->get_permissions()=3) echo 'Welcome to the Admin Panel'; else echo 'Sorry, You do not have access to this page'; ?> And the code that I'm actually trying to call with the if statement is: public function get_permissions() { $username = $_SESSION['admin_login']; global $db; $info = $db->get_row("SELECT `permissions` FROM `user` WHERE `username` = '" . $db->escape($username) . "'"); if(is_object($info)) return $info->permissions; else return ''; } This should be a simple way to call my pages that the user is authorized for by using an else if statement. Or So I thought

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  • Syntax problem when checking for a file

    - by Luke
    This piece of code has previously worked but I have C&P it to a new place and for some reason, it now won't work! <? $user_image = '/images/users/' . $_SESSION['id'] . 'a.jpg'; if (file_exists(realpath(dirname(__FILE__) . $user_image))) { echo '<img src="'.$user_image.'" alt="" />'; } else { echo '<img src="/images/users/small.jpg" alt="" />'; } ?> As you can see, I am checking for a file, if exists, showing it, if not, showing a placeholder. The $_SESSION['id'] variable does exist and is being used elsewhere within the script. Any ideas what the problem is? Thanks

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  • Windows 7 PATH not expanding

    - by trinithis
    I am using the following to create and edit environment variables for Windows 7. Control Panel\All Control Panel Items\System -> Advanced system settings -> Environment Variables Under System variables I have the following pertinant variables: PROG32=C:\Program Files (x86) REALDWG_SDK_DIR=%PROG32%\Autodesk\RealDWG 2011 Path=%REALDWG_SDK_DIR%;%PROG32%\Haskell\bin However, the following happens: C:\>echo %PROG32% C:\Program Files (x86) C:\>echo %Path% %REALDWG_SDK_DIR%;C:\Program Files (x86)\Haskell\bin Is it possible to have a chain of variables expand? If I rename Path to something else, I sometimes get the problem, and sometimes I don't.

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  • Which code snipplets in PHP can create a input form, which creates a new set of data in my mysql dat

    - by smiyazaki
    I am using the Google Maps API with parts in javascript and others in PHP. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> Google Maps AJAX + mySQL/PHP Example // var iconBlue = new GIcon(); iconBlue.image = 'icon.png'; iconBlue.shadow = ''; iconBlue.iconSize = new GSize(19, 19); iconBlue.shadowSize = new GSize(22, 20); iconBlue.iconAnchor = new GPoint(6, 20); iconBlue.infoWindowAnchor = new GPoint(5, 1); var iconRed = new GIcon(); iconRed.image = 'icon.png'; iconRed.shadow = ''; iconRed.iconSize = new GSize(19, 19); iconRed.shadowSize = new GSize(22, 20); iconRed.iconAnchor = new GPoint(6, 20); iconRed.infoWindowAnchor = new GPoint(5, 1); var customIcons = []; customIcons["restaurant"] = iconBlue; customIcons["bar"] = iconRed; function load() { if (GBrowserIsCompatible()) { var map = new GMap2(document.getElementById("map")); map.setMapType(G_SATELLITE_MAP); map.addControl(new GSmallMapControl()); map.addControl(new GMapTypeControl()); map.setCenter(new GLatLng(47.614495, -122.341861), 13); // Change this depending on the name of your PHP file GDownloadUrl("phpsqlajax_genxml.php", function(data) { var xml = GXml.parse(data); var markers = xml.documentElement.getElementsByTagName("marker"); for (var i = 0; i < markers.length; i++) { var name = markers[i].getAttribute("name"); var address = markers[i].getAttribute("address"); var type = markers[i].getAttribute("type"); var point = new GLatLng(parseFloat(markers[i].getAttribute("lat")), parseFloat(markers[i].getAttribute("lng"))); var marker = createMarker(point, name, address, type); map.addOverlay(marker); } }); } } function createMarker(point, name, address, type) { var marker = new GMarker(point, customIcons[type]); var html = "<b>" + name + "</b> <br/>" + address; GEvent.addListener(marker, 'click', function() { marker.openInfoWindowHtml(html); }); return marker; } //]]> (I suppose the php will be called by "GDownloadUrl("phpsqlajax_genxml.php", function(data) { ..." in the javascript part of the sourcecode of phpsqlajax_map.htm) Now I need another php file and the code snipplets for it, which creates an input form where I can add some new locations to the google map. Following code is used to create the xml file here: http://detektors.de/maptest/phpsqlajax_genxml.php The next step would be, trying to make an plugin for wordpress that I could easily post a blog entry with a new location on the same map, which displays already some other locations stored in the mysql database. thanks! <?php require("phpsqlajax_dbinfo.php"); function parseToXML($htmlStr) { $xmlStr=str_replace('<','<',$htmlStr); $xmlStr=str_replace('','>',$xmlStr); $xmlStr=str_replace('"','"',$xmlStr); $xmlStr=str_replace("'",''',$xmlStr); $xmlStr=str_replace("&",'&',$xmlStr); return $xmlStr; } // Opens a connection to a MySQL server $connection=mysql_connect ($host, $username, $password); if (!$connection) { die('Not connected : ' . mysql_error()); } // Set the active MySQL database $db_selected = mysql_select_db($database, $connection); if (!$db_selected) { die ('Can\'t use db : ' . mysql_error()); } // Select all the rows in the markers table $query = "SELECT * FROM markers WHERE 1"; $result = mysql_query($query); if (!$result) { die('Invalid query: ' . mysql_error()); } header("Content-type: text/xml"); // Start XML file, echo parent node echo ''; // Iterate through the rows, printing XML nodes for each while ($row = @mysql_fetch_assoc($result)){ // ADD TO XML DOCUMENT NODE echo ''; } // End XML file echo ''; ?

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  • checking the return code using python (MAC)

    - by cyberbemon
    I have written a script that checks if an SVN Repo is up and running, the result is based on the return value. import subprocess url = " validurl" def check_svn_status(): subprocess.call(['svn info'+url],shell=True) def get_status(): subprocess.call('echo $?',shell=True) def main(): check_svn_status() get_status() if __name__ == '__main__': main() The problem I'm facing is that if I change the url to something that does't exist I still get the return value as 0, but if I were to run this outside the script, i.e go to the terminal type svn info wrong url and then do a echo $? I get a return value of 1. But I can't re-create this in the python. Any guidelines ?

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  • how to make PHP lists all Linux Users?

    - by Data-Base
    Hello I want to build a php based site that (automate) some commands on my Ubuntu Server first thing I did was going to the file (sudoers) and add the user www-data so I can execute php commands with root privileges! # running the web apps with root power!!! www-data ALL=(ALL) NOPASSWD: ALL then my PHP code was <?php $command = "cat /etc/passwd | cut -d\":\" -f1"; echo 'running the command: <b>'.$command."</b><br />"; echo exec($command); ?> it returns only one user (the last user) !!! how to make it return all users? thank you

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  • PHP and Javascript: Getting file contents with Javascript variable

    - by celliott1997
    Could someone help me to understand why this isn't working? var uname = "<?php echo strtolower($_GET['un']) ?>"; var source = "<?php echo file_get_contents('accounts/"+uname+"') ?>"; console.log(source); I've been trying for a while to get this working and it just doesn't seem to. Before I added in the source variable, it was working fine and displayed the un variable on the page. Thanks.

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  • foreach invalid argument supplied and mysql fetch array issue

    - by La Myse
    i have this code which i use to print some fields from the database. My problem is that i get this error about foreach invalid argument supplied and a mysql fetch array problem. The code is this: foreach( $checked1 as $key => $value){ echo "<th> $value </th>"; } echo "</tr></thead>"; while($row = mysql_fetch_array($result)){ Where $checked1 is an array $checked1 = $_POST['checkbox']; What's the problem here? Thanks..

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  • display items vertically

    - by shawn swanson
    This is my code: <?php while($fetch_cat = mysql_fetch_array($rescat)) { $per_cnt++; ?> <li style="margin-left:10px;"> <a href="sub_cat.php?cat_id=<?php echo $fetch_cat['cat_id'];?>" style="color:#431603;text-decoration:none;"><?php echo stripslashes($fetch_cat['category_name']);?> </a> </li> <?php } ?> This is the output I am getting- alphabetical order horizontally: A B C D E F G H I J This is what I want to show- alphabetical order vertically: A E H B F I C G J D Please help. Thanks

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  • PHP_AUTH_USER only known in certain frames

    - by Rob
    Getting very confused by PHP_AUTH_USER. Within my web pages I have .htaccess files in every directory, controlling who can (and cant) see certain folders. In order to further customise the pages I was hoping to use PHP_AUTH_USER within the PHP code, i.e. tailor page contents based on the user. This only seems to work partially. The code snippets below hopefully demonstrate my problems. The main index.php creates a framed page with a menu structure in the top left hand corners, some irrelvant stuff in top right and then the tailor made contents in bottom frame. In top left the user is correctly shown, but in the bottom frame PHP_AUTH_USER doesnt seem to be set anymore (it returns empty and when printing all $HTTP_SERVER_VARS its not listed). Script.php is in a different path, but they all have .htaccess files in them and all other contents is displayed correctly. Why does it not know about PHP_AUTH_USER there? Running version php version 5.2.12 on chrome. index.php <FRAMESET ROWS="35%, *"> <FRAMESET COLS="25%, *"> <FRAME SRC="Menu.php"> <FRAME SRC="Something.php"> </FRAMESET> <FRAME SRC="../OtherPath/Script.php?large=1" name="outputlisting"> </FRAMESET> </FRAMESET> Menu.php <ul> <li>Reporting <ul> <li>Link1 <a href="../OtherPath/Script.php" target="outputlisting">All</a>, <a href="../OtherPath/Script.php?large=1" target="outputlisting">Big</a> </ul> <?php echo 'IP Address: ' . $_SERVER['REMOTE_ADDR'] . '<br />'; echo 'User: ' . $_SERVER['PHP_AUTH_USER']; ?> Script.php <?php echo 'User: ' . $_SERVER['PHP_AUTH_USER']; ?>

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  • Access a static variable by $var::$reference

    - by chuckg
    I am trying to access a static variable within a class by using a variable class name. I'm aware that in order to access a function within the class, you use call_user_func(): class foo { function bar() { echo 'hi'; } } $class = "foo"; all_user_func(array($class, 'bar')); // prints hi However, this does not work when trying to access a static variable within the class: class foo { public static $bar = 'hi'; } $class = "foo"; call_user_func(array($class, 'bar')); // nothing echo $foo::$bar; // invalid How do I get at this variable? Is it even possible? I have a bad feeling this is only available in PHP 5.3 going forward and I'm running PHP 5.2.6. Thanks.

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  • How to send mail using PHP?

    - by phpaddict
    I'm using Windows Vista OS. PHP, MySQL as the database and Apache web server. I want to send notification to those who want to join in my site. But the problem is when I click submit. It doesn't send anything to the email address of the user. What to do you think is the best solution for this? <?php $to = "[email protected]"; $subject = "Hi!"; $body = "Hi,\n\nHow are you?"; if (mail($to, $subject, $body)) { echo("<p>Message successfully sent!</p>"); } else { echo("<p>Message delivery failed...</p>"); } ?>

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