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  • PHP loop position

    - by Jordan Pagaduan
    Can someone help me on this. I'm made an image uploader and i want the image to make another tr if it reach to 5 pics so it will not overflow. Here is my code: Can someone help me on this. I'm made an image uploader and i want the image to make another tr if it reach to 5 pics so it will not overflow. Here is my code: $dbc = mysql_connect("localhost" , "root" , "") or die (mysql_error()); mysql_select_db('blog_data') or die (mysql_error()); $sql = "SELECT * FROM img_uploaded"; $result = mysql_query($sql); while($rows=mysql_fetch_array($result)) { if ($rows) { echo "<tr><td><img src='user_images/".$rows['img_name'] . "' width='100' height='100'></td></tr>"; } else { echo "<td><img src='user_images/".$rows['img_name'] . "' width='100' height='100'></td>"; } }

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  • PHP preg_match Math Function

    - by Matt
    I'm writing a script that will allow a user to input a string that is a math statement, to then be evaluated. I however have hit a roadblock. I cannot figure out how, using preg_match, to dissallow statements that have variables in them. Using this, $calc = create_function("", "return (" . $string . ");" ); $calc();, allows users to input a string that will be evaluated, but it crashes whenever something like echo 'foo'; is put in place of the variable $string.

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  • php won't detect all spaces in a string...

    - by user296516
    Hi guys, I've got a string that comes from a POST form where I want to replace all spaced with some other character. Here's that I did: $cdata = str_replace(" ","#",$cdata); And I got this. --- Contact-ID#=#148 [10274da8]#Sinhronizacija#=#private [1000137d]#Uzvards#=#Zom [1000137c]#Vards#=#Tana [1000130e]#Talrunis#=#3333 [1000130e]#Mobilais#=#5555 As you can see, spaced before "[10..." are still there. Any ideas what could be the problem?

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  • How to reformat date in PHP?

    - by Lisa
    I am pulling the dates of various posts from a database. The dates are in the following format: 2009-08-12 Numeric Year - Numeric Month - Numeric Day How can I reformat these dates to something more user friendly like: August 12, 2009 Numeric Month Numeric Date, Numeric Year Assuming that the date gotten from the mysql database is stored in a variable called: $date = $row['date_selected'];

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  • How to display XML in HTML in PHP?

    - by tomaszs
    Hello, I have a string with XML: $string = " <shoes> <shoe> <shouename>Shoue</shouename> </shoe> </shoes> "; And would like display it on my website like this: This is XML string content: <shoes> <shoe> <shouename>Shoue</shouename> </shoe> </shoes> So I would like to do it: on site, not in textbox without external libraries, frameworks etc. formatted with proper new lines formatted with tabs without colors etc., only text So how to do it in plain and simple way?

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  • PHP mail script html formatting

    - by Jimbly2
    someone kindly posted this code for me but it only returns in the resulting email - any ideas? Does it need a closing html tag? $mailHeader .= "Content-type: text/html; charset=iso-8859-1\r\n"; $formcontent ="<table border='1'>"; foreach ($_POST as $field=>$value) { $formcontent.="<tr>"; $formcontent .= "<td>$field:</td> <td>$value</td>"; $formcontent.="</tr>"; } $formcontent .= '<tr><td>User-Agent: </td><td>'.$_SERVER['HTTP_USER_AGENT'].'</td>'; $formcontent ="</table>"; Thanks, JIm

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  • Simple PHP query question: LIKE

    - by pg
    When I replace $ordering = "apples, bananas, cranberries, grapes"; with $ordering = "apples, bananas, grapes"; I no longer want cranberries to be returned by my query, which I've written out like this: $query = "SELECT * from dbname where FruitName LIKE '$ordering'"; Of Course this doesn't work, because I used LIKE wrong. I've read through various manuals that describe how to use LIKE and it doesn't quite make sense to me. If I change the end of the db to "LIKE "apples"" that works for limiting it to just apples. Do I have to explode the ordering on the ", " or is there a way to do this in the query?

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  • PHP uploads and checking

    - by user147685
    Hi all, I want to update/insert file by using upload box to the database but before that it will check for the file type thats only pdf can be upload. Somthing wrong with the codes and i dont know what..Please help here are part of my updates codes: $id=$rs['id']; $qry = "SELECT a.faillampiran FROM {$CFG->prefix}ptk_lampiran a, {$CFG->prefix}ptk b WHERE a.ptkid=$id and b.id=$id"; $sql = get_records_sql($qry); if($_POST['check']){ $ext = pathinfo($faillampiran,PATHINFO_EXTENSION); $err = "Upload Only PDF File. "; if ($ext =='pdf'){ $qry="UPDATE {$CFG->prefix}ptk_lampiran SET faillampiran='".$faillampiran."' WHERE ptkid='".$_GET['id']."'"; $sql=mysql_query($qry); $qry = "SELECT a.faillampiran FROM {$CFG->prefix}ptk_lampiran a, {$CFG->prefix}ptk b WHERE b.id = '".$rs[id]."' AND a.ptkid = '".$rs[id]."' "; $sql = get_records_sql($qry); foreach($sql as $rs){ ?> <?=basename($rs->faillampiran); ?><br> <? } ?> <tr><td></td><td></td><td> <input type="file" size="50" name="faillampiran" alt="faillampiran" value= "<?=$faillampiran;?>" /> <input type="submit" name="edit" value="Muat naik fail ini" /><br /> </td></tr> } else { echo "<script>alert('$err')</script>"; } } else { foreach($sql as $rs){ ?> <?=basename($rs->faillampiran); ?><br> <? } ?> <input type="file" size="50" name="faillampiran" alt="faillampiran" value= "<?=$faillampiran;?>" /> <input type="submit" name="edit" value="Muat naik fail ini" /><br /> <?} Sori bout the messiness of the codes, im still a beginner in this languages. thx

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  • MySQL/PHP Search Efficiency

    - by iMaster
    Hi! I'm trying to create a small search for my site. I've tried using full-text index search, but I could never get it to work. Here is what I've come up with: if(isset($_GET['search'])) { $search = str_replace('-', ' ', $_GET['search']); $result = array(); $titles = mysql_query("SELECT title FROM Entries WHERE title LIKE '%$search%'"); while($row = mysql_fetch_assoc($titles)) { $result[] = $row['title']; } $tags = mysql_query("SELECT title FROM Entries WHERE tags LIKE '%$search%'"); while($row = mysql_fetch_assoc($tags)) { $result[] = $row['title']; } $text = mysql_query("SELECT title FROM Entries WHERE entry LIKE '%$search%'"); while($row = mysql_fetch_assoc($text)) { $result[] = $row['title']; } $result = array_unique($result); } So basically, it searches through all the titles, body-text, and tags of all the entries in the DB. This works decently well, but I'm just wondering how efficient would it be? This would only be for a small blog, too. Either way I'm just wondering if this could be made any more efficient.

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  • php mysql , update query is not working

    - by Michael
    I'm trying to update some info into database but for some reasons it doesn't update. Also I'm not getting error in the server logs. mysql_query("UPDATE `view_item` SET `item_number` = $item_number, `title` = $title, `price` = $price, `shipping` = $shipping, `location` = $location, `start_time` = $start_time, `end_time` = $end_time, `seller_userName` = $seller_userName, `seller_UserNum` = $seller_UserNum, `number_of_bids` = $number_of_bids, `picture_link` = $picture_link WHERE `item_number` = $item_number");

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  • PHP Session Array Value keeps showing as "Array"

    - by Nerathas
    Hello, When sending data from a form to a second page, the value of the session is always with the name "Array" insteed of the expected number. The data should get displayed in a table, but insteed of example 1, 2, 3 , 4 i get : Array, Array, Array. (A 2-Dimensional Table is used) Is the following code below a proper way to "call" upon the stored values on the 2nd page from the array ? $test1 = $_SESSION["table"][0]; $test2 = $_SESSION["table"][1]; $test3 = $_SESSION["table"][2]; $test4 = $_SESSION["table"][3]; $test5 = $_SESSION["table"][4]; What exactly is this, and how can i fix this? Is it some sort of override that needs to happen? Best Regards.

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  • How can I "flip" an image using PHP?

    - by learner
    Here's what I've tried: $image = "images/20100609124341Chrysanthemum.jpg"; $degrees = 40; // Content type header('Content-type: image/jpeg'); // Load $source = imagecreatefromjpeg($filename); // Rotate $rotate = imagerotate($source, $degrees, 0); // Output imagejpeg($rotate); ...But I get no output. Can anyone tell me what's wrong with this?

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  • PHP turn off errors - in one file only

    - by Industrial
    Hi! I am well aware about error_reporting(0); & ini_set('display_errors', "Off"); to make error messages go away. What would be an appropriate way to do this - for a specific file or part of code only? Surpressing errors with @'s seems like a bad idea since it apparently slows the code down... Thanks!

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  • Add 30 seconds to the time with PHP

    - by Sam
    Hey all, just wondering how I can add 30 seconds on to this? $time = date("m/d/Y h:i:s a", time()); Thankyou, again. I wasn't sure how to do it because it is showing lots of different units of time, when I only want to add 30 seconds.

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  • How to create PHP method linking?

    - by Kerry
    I've seen other objects that do this: $obj->method1()->method2(); How do I do that? Is each function just modifying the pointer of an object or returning a pointer? I don't know the proper term for this style -- if anyone could help me with that, it would be great.

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  • Php random row help...

    - by Skillman
    I've created some code that will return a random row, (well, all the rows in a random order) But i'm assuming its VERY uneffiecent and is gonna be a problem in a big database... Anyone know of a better way? Here is my current code: $count3 = 1; $count4 = 1; //Civilian stuff... $query = ("SELECT * FROM `*Table Name*` ORDER BY `Id` ASC"); $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $count = $count + 1; $civilianid = $row['Id']; $arrayofids[$count] = $civilianid; //echo $arrayofids[$count]; } while($alldone != true) { $randomnum = (rand()%$count) + 1; //echo $randomnum . "<br>"; //echo $arrayofids[$randomnum] . "<br>"; $currentuserid = $arrayofids[$randomnum]; $count3 += 1; while($count4 < $count3) { $count4 += 1; $currentarrayid = $listdone[$count4]; //echo "<b>" . $currentarrayid . ":" . $currentuserid . "</b> "; if ($currentarrayid == $currentuserid){ $found = true; //echo " '" .$found. "' "; } } if ($found == true) { //Reset array/variables... $count4 = 1; $found = false; } else { $listdone[$count3] = $currentuserid; //echo "<u>" . $count3 .";". $listdone[$count3] . "</u> "; $query = ("SELECT * FROM `*Tablesname*` WHERE Id = '$currentuserid'"); $result = mysql_query($query); $row = mysql_fetch_array($result); $username = $row['Username']; echo $username . "<br>"; $count4 = 1; $amountdone += 1; if ($amountdone == $count) { //$count $alldone = true; } } } Basically it will loop until its gets an id (randomly) that hasnt been chosen yet. -So the last username could take hours :P Is this 'bad' code? :P :(

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  • Regular expression matching in php

    - by user1836428
    I have this regexp: /(.*)(([0-9]([^a-zA-Z])*){7,}[0-9])(.*)/. Given the following values 0654535263 065453-.-5263 065asd4535263 Expected Results 06**** 06**** 06**** Actual Results 0654535263 06**** 065asd4535263 It does not match the last row because of the letters (I want to match from 0-3 letters) and it matches only last occurence (in the second row in example, it skips first row).

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