|
PHP - How Can I Use A Variable To Get An Array From Another Php File?
I am trying to use the variable $RATETHESE to include an array into file A. The array is in file B.
PhP File A include_once(dirname(__FILE__) . '/PhpFileB.php'); global $RATETHESE; $VALID_RATING_IDS = $RATETHESE; Php File B $RATETHESE = "array("1", "2");" For some reason it is printing the array on the page, and the array is not being used to run the script. Did i do anything wrong? Similar TutorialsHi, I want to store files to variable array using glob() like $files[0] = xyz.txt $files[1] = pqr.txt . . . $files[n] = nfile.txt I know how to list files from directory using glob() Code: [Select] <?php foreach (glob("*.txt") as $filename) { echo "$filename size " . filesize($filename) . "\n"; } ?> How can I do that ? Hi guys im in the middle of optimizing code.. Code: [Select] $sql = "SELECT * FROM sa_enemystats WHERE username='$username'"; $res = mysql_query($sql); while ($row = mysql_fetch_assoc($res)) { $enemy['current_health'] = $row['current_health']; $enemy['current_skill'] = $row['current_skill']; $enemy['level'] = $row['level']; $enemy['damage'] = $row['damage']; $enemy['evade'] = $row['evade']; $enemy['accuracy'] = $row['accuracy']; $enemy['speed'] = $row['speed']; $enemy['luck'] = $row['luck']; echo 'debug: variables synced with db table'; }I know that the setting of these variables are messy and can be done in a better way... but how? I have tried foreach and copying other's code for an array copy but it lead no where. Then again my syntax could be wrong... I used something like Code: [Select] foreach ( $enemy[$value] as $row => $value) { $enemy[$value] = $row[$value]; } I need a help in the following : I have an admin page from where the admin attaches an gif image.This attached image should be shown to the user as a scrolling image. The code for the scrolling image is done through javascript. So my actual problem is getting the name of the attached gif image to the javascript array which is used for scrolling horizontally. hi all, I have an language pack for example: languages/en.php: Code: [Select] $en['mail']['letter closing'] = "regards,\n your friend!"; and in my config: Code: [Select] $language = "en"; $include_language = @include("languages/".$language.".php"); if(!($include_language)) { $try_default_language = @include("languages/nl.php"); if(!($try_default_language)) { echo "kan de taalpakket niet vinden<br>"; echo "Could not find the language pack.<br>"; echo "example on error: ".$test." shows nothing"; exit; } } In my function I want to include the language pack for example i have $language = 'en' so I want to include $en['general']['letter closing'] I will do this: Code: [Select] global $language,${$language}['general']; But that gives an error unexpected '[' blah blah. How can i call the variable variable array in the valid php way? Code: [Select] $word = 'numbers'; $numbers= array('1', '2', '3', '4'); echo $$word[0]; I expected the output to be '1'. It ended up being nothing... Why does this not work? Is it not possible to have a variable variable array? And if not, is there a workaround? Cheers, Joe I'm new to php and i have a html form with 2 forms and 2 submit buttons, 1 of the forms is solely for uploading an image i downloaded the php script as i don't understand the code well enough to write it myself. The second form asks for your name and email then when you click submit it writes a new html file with the included information. I have managed to upload a file successfully and write the html document with the information successfully but i can't get the image to write into the html document. My file_upload.php script contains the variable for the image name, now how would i get the variable string from file_upload.php and use that name in compile.php. Here's my code. -------------- Form.html -------------- <!DOCTYPE html> <html lang="en"> <head> <title></title> </head> <body> <form action="compile.php" method="post"> <p>input name:<input type="text" name="name"/></p><br /> <p>email:<input type="text" name="email" /></p><br /> </form> <br /> <form enctype="multipart/form-data" method="post" action="file_upload.php"> Choose your file he <input name="file1" type="file" /><br /><br /> <input type="submit" value="Upload It" /> </form> </body> </html> ------------- compile.php ------------- <?php $name = $_POST['name']; $price = $_POST['price']; $desc = $_POST['desc']; $file = "new.html"; $handle = fopen($file,'w'); $data = "<!DOCTYPE html> <html> <head> <title></title> </head> <body> <table style='border-radius:8px;'> <tr> <td style='border-radius:8px;'> <h1>$name</h1> </td> </tr> </table> <table> <tr> <td> <ul> <lh>Product name: $name</lh> <br><br> <li>Price: $price</li><br> <li>Description: $desc</li><br> </ul> </td> <td> <img src='uploads/$fileName' width='' height='' alt='' title='' /> </td> </tr> </table> </body> </html>"; fwrite($handle, $data); print "data written"; fclose($handle); ?> ------------ file_upload.php ------------ <?php // Set local PHP vars from the POST vars sent from our form using the array // of data that the $_FILES global variable contains for this uploaded file $fileName = $_FILES["file1"]["name"]; // The file name $fileTmpLoc = $_FILES["file1"]["tmp_name"]; // File in the PHP tmp folder $fileType = $_FILES["file1"]["type"]; // The type of file it is $fileSize = $_FILES["file1"]["size"]; // File size in bytes $fileErrorMsg = $_FILES["file1"]["error"]; // 0 for false... and 1 for true // Specific Error Handling if you need to run error checking if (!$fileTmpLoc) { // if file not chosen echo "ERROR: Please browse for a file before clicking the upload button."; exit(); } else if($fileSize > 5000000000) { // if file is larger than we want to allow echo "ERROR: Your file was larger than 5mb in file size."; unlink($fileTmpLoc); exit(); } else if (!preg_match("/.(gif|jpg|png)$/i", $fileName) ) { // This condition is only if you wish to allow uploading of specific file types echo "ERROR: Your image was not .gif, .jpg, or .png."; unlink($fileTmpLoc); exit(); } // Place it into your "uploads" folder mow using the move_uploaded_file() function move_uploaded_file($fileTmpLoc, "uploads/$fileName"); // Check to make sure the uploaded file is in place where you want it if (!file_exists("uploads/$fileName")) { echo "ERROR: File not uploaded<br /><br />"; echo "Check folder permissions on the target uploads folder is 0755 or looser.<br /><br />"; echo "Check that your php.ini settings are set to allow over 2 MB files, they are 2MB by default."; exit(); } // Display things to the page so you can see what is happening for testing purposes echo "The file named <strong>$fileName</strong> uploaded successfuly.<br /><br />"; echo "It is <strong>$fileSize</strong> bytes in size.<br /><br />"; echo "It is a <strong>$fileType</strong> type of file.<br /><br />"; echo "The Error Message output for this upload is: <br />$fileErrorMsg"; ?> Is this right? $admin = array('24.68.200.235','24.68.214.97'); $visitor_ip = $_SERVER['REMOTE_ADDR']; if($visitor_ip == $admin) { echo "k"; } else { die("dead"); Hello. Can someone show the syntax to echo the variable from the array below that is showing "2010-08-31 23:19"? I need to grab that date and use it elsewhere in my script. Thanks. Code: [Select] array(19) { ["name"]=> string(8) "com_jumi" ["params"]=> array(0) { } ["mainframe"]=> &object(JSite)#2 (6) { ["_clientId"]=> int(0) ["_messageQueue"]=> array(0) { } ["_name"]=> string(4) "site" ["scope"]=> string(8) "com_jumi" ["_errors"]=> array(0) { } ["requestTime"]=> string(16) "2010-08-31 23:19" } Hey, im struggeling with a small piece of code.... Thats my code: Global $user_ID; $user_ID = get_current_user_id(); echo"User number $user_ID is loggedin";
Echo gives me this:
Now i want to add it to my array: $atts['href'] .= $user_ID . 'abc'; 'href' is set to "https://test.com/" What im getting now ist https://test.com/abc
Can someone tell my why the '2' is missing??
hello all, I have an array value in my post array ($_POST['check0'] ) that I am trying to break up into php variables. For some reason if I do this: $var1 = $_POST['check0'][0]; all I get is the first letter of the string from the first value of the array? Hi, I am basically trying to apply a different amount to $defineMe for certain numbers of an array .. in this case it's array numbers 11 - 16 Code: [Select] <?php if ($defineNumber > 11 and $defineNumber < 16) { $defineMe = 35.00; } else { $defineMe = 50.00; } $defineNumber[01] = array('name'=>'Product 1'); $defineNumber[02] = array('name'=>'Product 2'); $defineNumber[03] = array('name'=>'Product 3'); $defineNumber[04] = array('name'=>'Product 4'); $defineNumber[05] = array('name'=>'Product 5'); $defineNumber[06] = array('name'=>'Product 6'); $defineNumber[11] = array('name'=>'Product 11'); $defineNumber[12] = array('name'=>'Product 12'); $defineNumber[13] = array('name'=>'Product 13'); $defineNumber[14] = array('name'=>'Product 14'); $defineNumber[15] = array('name'=>'Product 15'); $defineNumber[16] = array('name'=>'Product 16'); ?> Hi Everyone
Please checkout the below code:
foreach ($joiners as $joiner) { $town = explode(',', $joiner['address']); if (isset($town[2])) { $town = $town[2]; $get_available_towns[] = $town; $get_available_towns = array(); }
foreach ($joiners as $joiner) {
$town = explode(',', $joiner['address']);
if (isset($town[2])) {
$town = $town[2];
$get_available_towns[] = $town;
$get_available_towns = array();
}
I don't really understand what is happening. I know the code and everything is correct and it works but I really don't understand how the loop, arrray and variable is working. I wanted to print_r the $get_available_towns outside the loop but it would not work without the square brackets but also I did not even know the square brackets were required. The point was to loop through the data and print all the values. I understand how loops, array's and variables work but I've not seen them used like this. This makes me think I am missing some basic knowledge but I have looked around and I can't see any examples of how loops and arrays are used like this. I have also checked the php manual but still no joy. Can you guys tell me where I could find the information that would have examples on how something like the above code would work?
Hi, I need help with this, I am quite new to arrays. This one works fine: echo array_sum(array(34,19,22,17,7,22,8,23,17,31,21,32,7,29,13,7))/count(array(34,19,22,17,7,22,8,23,17,31,21,32,7,29,13,7)); which gives "19..." it gets average number. What I want to put the values in the array in a variable so I do this: $t = "34,19,22,17,7,22,8,23,17,31,21,32,7,29,13,7"; so why can't I use array like this: echo array_sum(array($t))/count(array($t)); I need to use variable in array, please someone help. Hello Guys. I have a question here.. I have a photo upload script that uploads photos to the server and then sends the photo name to another page via POST. I only have one var name called $image_all when I echo it this is what I get (When uploading 2 files) Code: [Select] This is the list 2011-12-05_0751.png,2011-12-05_0747.png I also tested with "explode" Code: [Select] print_r(explode(',', $image_all, 10)); and I get this Code: [Select] Array ( [0] => 2011-12-05_0751.png [1] => 2011-12-05_0747.png [2] => What I would like for it to do is assign an incremental var name to each image uploaded and passed in the POST. Like this $image1 = 2011-xx.png $image2 = 2011-xx.png $image3 = 2011-xx.png It should created the variable names dynamically based on the number of images (Up to 10) Thanks for your help! Hi, I have a variable named $siteName stored in the server.php file in the data directory data/server.php Now in the header.php page I want to retrieve the $siteName variable so it can be used in the header information. This is what I have but it isnt working. server.php Code: [Select] <?php $siteName = 'My+Test+Site'; ?><?php $adminEmail = 'pwithers2009@hotmail.co.uk'; ?><?php $sendmailLoc = '/usr/sbin/sendmail'; ?><?php $imgdir = 'http://www.tropicsbay.co.uk/images/'; ?><?php $imgdirbase = '/home/tropicsb/public_html/images/'; ?> header.php Code: [Select] <?php include("data/server.php"); $siteName = $_GET['siteName']; echo ' <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>$siteName Admin Panel</title> <link rel="stylesheet" href="admin.css" type="text/css" /> </head> <div id="pagehead"> <div id="navigation"> <img alt="" src="images/leftNav.gif" height="32" width="4" id="leftNav" /> <img alt="" src="images/rightNav.gif" height="32" width="4" id="rightNav" /> <div id="globalLink"> <a href="admin.php?cmd=manage&username=admin&password=$adminpw" id="gl1" class="glink" onmouseover="ehandler(event,menuitem1);">Users</a><a href="admin.php?cmd=editTemplate&username=admin&password=$adminpw" id="gl2" class="glink" onmouseover="ehandler(event,menuitem2);">Templates</a><a href="admin.php?cmd=mysqlBackup&username=admin&password=$adminpw" id="gl3" class="glink" onmouseover="ehandler(event,menuitem3);">Database</a><a href="admin.php?cmd=paymentLog&username=admin&password=$adminpw" id="gl4" class="glink" onmouseover="ehandler(event,menuitem4);">Payment</a><a href="admin.php?cmd=profileFields&username=admin&password=$adminpw" id="gl5" class="glink" onmouseover="ehandler(event,menuitem5);">Setup</a><a href="admin.php?cmd=changeAdminpw&username=admin&password=$adminpw" id="gl6" class="glink" onmouseover="ehandler(event,menuitem6);">Preferences</a><a href="admin.php?cmd=logout&username=admin&password=$adminpw" id="gl7" class="glink" onmouseover="ehandler(event,menuitem6);">Logout</a></div> </div></div> <div id="pagelayout"> <img alt="" src="images/leftCurve.gif" height="6" width="6" id="left" /> <img alt="" src="images/rightCurve.gif" height="6" width="6" id="right" /> <div id="pageName"> <h2>$siteName Admin Panel<h2> </div>'; ?> Any ideas on how to do this, it seems to work ok in one of my other scripts. Thanks I am trying to get an swf to upload an image along with user information. I was able to get the upload and variables send working individually, but together it stops with no visible errors. The following is a link: http://jarrodverhagen.com/upload/swfindex.htm to a tutorial that is using almost identical code to mine, and has been used by others, but still will not work on my site. Is there a reason supposedly correct code wouldn't work on my site? The original link for the tutorial is: http://www.developphp.com/Flash_tutoria ... sh_and_AS3 can i add php variables into a js file? i have tried this: document.location = <?php echo $_SESSION['fwd_page']; ?>; but it doesnt work. My catalog uses a template file to list product details. One part of it is responsible for putting these details into nifty little tabs: <?php $template = '{magictabs style=black_rounded, tabwidth=110px}'; $template .= 'Technical Specifications'; $template .= '::'; $template .= '<table colspan="3"><tr><td>'; $template .= $details; $template .= '</td></tr></table>'; $template .= '||||'; ?> I don't want it to output $details, I want it to output a separate script instead. How do I include this separate file as a variable? I want this: <?php include('specs.php'); ?> to replace the $details variable. Alright I am so close but can not figure this out, thanks in advance for any help.
So what I am trying to do is take the input variables from a form and have them load a specific PDF .
<html>
<body>
<?php
if( $_GET["startpoint"] || $_GET["endpoint"])
{
$a = $_GET["startpoint"];
$b = $_GET["endpoint"];
echo "Your Startpoint is:". $_GET["startpoint"]. "<br />";
echo "Your Endpoint is: ". $_GET["endpoint"]. "<br />";
echo $a,$b,".pdf";
}
?>
</body>
</html>
This is what I have and it works to display what the two points are what the final pdf file should be.
The final echo generates the PDF name
echo $a,$b,".pdf";How can I get it to load a file using this as a variable or the name it generates, thanks. when tyring to pass a variable to a extenal javascript file it just shows up as a sting and the variable isnt beening executed
var timezone_offset = <?= $timezone_offset ?>;im guessing its probably down to the fact the variable is being passed from my framework controller... so im wondering how do you send a variable to a external javascript file when using a framework?
$this->view->timezone_offset = "test"; // in my controller
// passing varibales to template in my view
if (preg_match("/\.html$/i", $file_name))
{
extract($this->_variables);
require_once PRIVATE_DIRECTORY . 'application' . DS . 'views' . DS . $file_name;
}
thanks guys
|
|||||||