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  • How to know record has been updated successfully in php

    - by Lisa Ray
    $sql = "UPDATE...."; if(mysql_query($sql)) { $_SESSION['Trans']="COMMIT"; header("location:result.php"); exit; } else { $_SESSION['Trans']="FAIL"; $_SESSION['errors'] = "Error: Sorry! We are unable to update your Profile, Please contact to PNP HelpDesk."; header("location:result.php"); exit; }//end IF data is getting updated then why compiler is not coming inside IF condition.

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  • PHP how to send email basic

    - by user329394
    hi all, im using PHP tu send email from my localhost to other people. can tell me what should i need to do that? for eg should i need to install mailserver? If im not mistaken theres is a language that u dont need a mailsever to send email. is it right? inside the PHP.ini, there are [mail function] how to configure this? i checked on the net, but not really understand how it works. [mail function] ; For Win32 only. SMTP = localhost smtp_port = 25 sendmail_from [email protected] //not sure how to write this?

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  • Ajax gets nothing back from the php.

    - by ShaMun
    Jquery i dont have alert and firefox i dont have anything in return. The code was working before, database query have successfull records also. What i am missing??? Jquery ajax. $.ajax({ type : "POST", url : "include/add_edit_del.php?model=teksten_display", data : "oper=search&ids=" + _id , dataType: "json", success : function(msg){ alert(msg); } }); PHP case 'teksten_display': $id = $_REQUEST['ids']; $res = $_dclass-_query_sql( "select a,b,id,wat,c,d from tb1 where id='" . $id . "'" ); $_rows = array(); while ( $rows = mysql_fetch_array ($res) ) { $_rows = $rows; } //header('Cache-Control: no-cache, must-revalidate'); //header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); header('Content-type: application/json'); echo utf8_encode( json_encode($_rows) ) ; //echo json_encode($_rows); //var_dump($_rows); //print_r ($res); break; Firefox response/request header Date Sat, 24 Apr 2010 22:34:55 GMT Server Apache/2.2.3 (CentOS) X-Powered-By PHP/5.1.6 Expires Thu, 19 Nov 1981 08:52:00 GMT Cache-Control no-store, no-cache, must-revalidate, post-check=0, pre-check=0 Pragma no-cache Content-Length 0 Connection close Content-Type application/json Host www.xxxx.be User-Agent Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.9.1.9) Gecko/20100330 Fedora/3.5.9-2.fc12 Firefox/3.5.9 Accept application/json, text/javascript, */* Accept-Language en-us,en;q=0.5 Accept-Encoding gzip,deflate Accept-Charset ISO-8859-1,utf-8;q=0.7,*;q=0.7 Keep-Alive 300 Connection keep-alive Content-Type application/x-www-form-urlencoded; charset=UTF-8 X-Requested-With XMLHttpRequest Referer http://www.xxxx.be/xxxxx Content-Length 17 Cookie csdb=2; codb=5; csdbb=1; codca=1.4; csdca=3; PHPSESSID=benunvkpecqh3pmd8oep5b55t7; CAKEPHP=3t7hrlc89emvg1hfsc45gs2bl2

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  • retrieved upload images in php

    - by hunter
    i want to retrieve image from client (ipod) programmed in objective c i use the following code $TARGET_PATH = "pics/"; $image = $_FILES['photo']; $TARGET_PATH =$TARGET_PATH . basename( $_FILES['photo']['name']); $TARGET_PATH =$TARGET_PATH.".jpg"; if(file_exists($TARGET_PATH)) { $TARGET_PATH =$TARGET_PATH .uniqid() . ".jpg"; } if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { $TARGET_PATH="http://www.".$_SERVER["SERVER_NAME"]."/abc/".$TARGET_PATH; echo $TARGET_PATH; echo "image upload successfully";} else{ echo "could not upload image"; } this code upload five to six images successfully and after that it gives me error i.e Notice: Undefined index: photo in /home/abc/public_html/abc.com/fish/mycatch_post.php on line 42 Notice: Undefined index: photo in /home/abc/public_html/abc.com/fish/mycatch_post.php on line 53 could not upload image

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  • javascript not loading after output of PHP file_get_contents() string

    - by mives
    I have the following structure: -uploadpage.html -index.php -resources -jquery.1.4.2.min.js On my index.php, I get the contents then output uploadpage.html through this code: $output = file_get_contents("uploadpage.html"); echo $output; My upload.html has the javascript at the bottom: <script type="text/javscript" src="./resources/jquery-1.4.2.min.js"></script> <script type="text/javascript"> $(document).ready( function() { callSomething(); } ); function callSomething() {return true;} </script> However this results in an error "$ is not defined" which means that the jquery js file is not loaded. Any ideas?

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  • Create table with PHP and populate from MySQL

    - by typoknig
    Hi all, I am creating a table to display on a web page and that table is populated from data in a MySQL database. I am trying to do a couple of things that are making it difficult for me. First I am trying to have call the PHP code that exists in a separate file in HTML via JavaScript. I think I have that working right but I am not 100% sure (because the table will not display). I think it is working right because some of the code for the table (which is in the PHP file) displays in FireBug. Second I am trying to make it so the rows alternate colors for easy viewing too. My PHP code so far is below. The table does not display at all in any browser. $query = "SELECT * FROM employees"; $result = mysql_query($query); $num = mysql_num_rows($result); echo '<table>'; for ($i = 0; $i < $num; $i++){ $row = mysql_fetch_array($result); $id = $row['id']; $l_name = $row['l_name']; $f_name = $row['f_name']; $ssn = $row['ssn']; $class = (($i % 2) == 0) ? "table_odd_row" : "table_even_row"; echo "<tr>"; echo "<td class=" . $class . ">$wrap_id</td>"; echo "<td class=" . $class . ">$wrap_l_name</td>"; echo "<td class=" . $class . ">$wrap_f_name</td>"; echo "<td class=" . $class . ">$wrap_ssn</td>"; echo "</tr>"; } echo '</table>'; mysql_close($link); }

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  • Guidance for Php for a beginner

    - by luckyluke
    I've just started to learn PHP. I found the $_POST variable is not working and posted the same at the below link $_POST[] not working in php and as per the advise i installed XAMPP. But still the proble of $_POST variable is not solved. Now i've a doubt whether i need to configure any global variable to make $_POST work. I'm totally lost on this and dont know how to proceed. Any help on this is verryy much appreciated. Thanks.

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  • How to check if a checkbox/ radio button is checked in php

    - by user225269
    I have this html code: <tr> <td><label><input type="text" name="id" class="DEPENDS ON info BEING student" id="example">ID</label></td> </tr> <tr> <td> <label> <input type="checkbox" name="yr" class="DEPENDS ON info BEING student"> Year</label> </td> </tr> But I don't have any idea on how do I check this checkboxes if they are checked using php, and then output the corresponding data based on the values that are checked. Please help, I'm thinking of something like this. But of course it won't work, because I don't know how to equate checkboxes in php if they are checked: <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $id = mysql_real_escape_string($_POST['idnum']); if($_POST['id'] == checked & $_POST['yr'] ==checked ){ $result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'"); echo "<table border='1'> <tr> <th>IDNO</th> <th>YEAR</th> </tr>"; while($row = mysql_fetch_array($result2)) { echo "<tr>"; echo "<td>" . $row['IDNO'] . "</td>"; echo "<td>" . $row['YEAR'] . "</td>"; echo "</tr>"; } echo "</table>"; } mysql_close($con); ?>

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  • Force freeing memory in PHP

    - by DBa
    Hi everybody, in a PHP program, I sequentially read a bunch of files (with file_get_contents), gzdecode them, json_decode the result, analyze the contents, throw the most of it away, and store about 1% in an array. Unfortunately, with each iteration (I traverse over an array containing the filenames), there seems to be some memory lost (according to memory_get_peak_usage, about 2-10 MB each time). I have double- and triplechecked my code, I am not storing unneded data in the loop (and the needed data hardly exceeds about 10MB overall), but I am frequently rewriting (actually, strings in an array). Apparently, PHP does not free the memory correctly, thus using more and more RAM until it hits the limit. Is there any way to do a forced garbage collection? Or, at least, to find out where the memory is used? Thanks in advance, Dmitri

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  • how to get the value of checksum in php header

    - by sumit
    I have a header in php which contains a link like <?php header("Location: "."https://abc.com/ppp/purchase.do?version=3.0&". "merchant_id=<23255>&merchant_site_id<21312>&total_amount=<69.99>&". "currency=<USD>&item_name_1=<incidentsupporttier1>&item_amount_1=<1>&". "time_stamp=<2010-06-14.14:34:33>&**checksum=<calculated_checksum>**"); ?> when i run this page the value of checksum is calculated and the link is opened now how checksum is calculated? calculated_checksum=md5(abc); md5 is an algorithm which calculates the value of checksum based on certain values inside the bracket. now i want to know how can i pass the value of checksum in the header url

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  • Help with php custom popup

    - by user329394
    Hi i have create a custom popup using javascript to use in PHP file. thw popup will appear when user click My Special Website i need to change the title header. Can help me how to do it?? //function: function myPopup() { window.open( "<?=$CFG->wwwroot.'/example.php';?>" ,"myWindow", " height = 300, width = 600, statusbar=0, scrollbars=0,resizable=0, location=0,status=0, directories=0, menubar =0, toolbar =0 , left = 262,top = 234" ) } // call: <a href="#" onclick="myPopup() " >My Special WebSite </a> <span class="style1">*</span> See my sample page:

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  • PHP - javascript validation radio button

    - by user1806136
    i have a form with 3 sets of radio buttons. i want to set a simple javascript validation alert to appear when user clicks on submit when one of the fields is null. how can i do that using javascript ? my code so far is .. <?php session_start(); $Load=$_SESSION['login_user']; include('../connect.php'); if (isset($_POST['submit'])) { $v1 = intval($_POST['v1']); $v2 = intval($_POST['v2']); $v3 = intval($_POST['v3']); $total = $v1 + $v2 + $v3 ; mysql_query("INSERT into Form1 (P1,P2,P3,TOTAL) values('$v1','$v2','$v3','$total')") or die(mysql_error()); header("Location: mark.php"); } <center><form method="post" action="mark.php" > <tr> <th > School Evaluation <font size="4" > </font></th> <tr> <th > Criteria <font size="4" > </font></th> <th> 4<font size="4" > </font></th> <th> 3<font size="4" > </font></th> <th> 2<font size="4" > </font></th> <th> 1<font size="4" > </font></th> </tr> <tr> <th> Your attendance<font size="4" > </font></th> <td> <input type="radio" name ="v1" value = "4" onclick="updateTotal();"/></td> <td> <input type="radio" name ="v1" value = "3" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v1" value = "2" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v1" value = "1" onclick="updateTotal();" /></td> </tr> <tr> <th > Your grades <font size="4" > </font></th> <td> <input type="radio" name ="v2" value = "4" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v2" value = "3" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v2" value = "2" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v2" value = "1" onclick="updateTotal();" /></td> </tr> <tr> <th >Your self-control <font size="4" > </font></th> <td> <input type="radio" name ="v3" value = "4" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v3" value = "3" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v3" value = "2" onclick="updateTotal();" /></td> <td> <input type="radio" name ="v3" value = "1" onclick="updateTotal();" /></td> </tr> </tr> </table> i have put <br> <td><input type="submit" name="submit" value="Submit" onClick="return validation(form);"> <input type="reset" name="clear" value="clear" style="width: 70px"></td> </form> i have try alot of codes but no alert appears!

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  • Storing an Image with php?

    - by Chris
    I'm trying to store an Image in my website so I can use it easily but I found this php code from here and I can't quite make much sense of it.. I'm just starting php and I dont quite know what to change and what to keep.. I'd greatly appreciate it if you could explain this a little better for me, thanks. <?php $allowedExts = array("jpg", "jpeg", "gif", "png"); $extension = end(explode(".", $_FILES["file"]["name"])); if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/png") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 20000) && in_array($extension, $allowedExts)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file"; } ?>

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  • Pretty-print HTML via PHP without validation?

    - by brianjcohen
    I'd like to automatically pretty-print (indentation, mostly) the HTML output that my PHP scripts generate. I've been messing with Tidy, but have found that in its efforts to validate and clean my code, Tidy is changing way too much. I know Tidy's intentions are good but I'm really just looking for an HTML beautifier. Is there a simpler library out there that can run in PHP and just do the pretty-printing? Or, is there a way to configure Tidy to skip all the validation stuff and just beautify?

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  • php var_dump returning blank array

    - by Oroku
    I'm following this tutorial on youtube: https://www.youtube.com/watch?v=VSF5p00uorc Around 4:28 - I get the following array when I load the sql.php page as in the tutorial: array(8) { [1]=> NULL [2]=> NULL [3]=> NULL [4]=> NULL [5]=> NULL [6]=> NULL [7]=> NULL [8]=> NULL } I do have 8 ids all with values in my data table. This is my code: <?php $con = mysqli_connect("localhost","root","","test2") or die('error'); $query = "select * from data"; $result = mysqli_query($con,$query); while($row = mysqli_fetch_array($result)) { $id = $row['id']; $feed = $row['feed']; $data[$id] = $feed; } var_dump($data); ?>

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