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  • Declaring variable as PHP class

    - by iamdadude
    I have a variable called $theclass and it's a string "Home_class". How can I define a class out of that string? The reason I need to do this is that the variable will change and I want to be able to declare the class that the variable is equal to. Is this possible? Thanks for any help.

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  • Display Yearly Report When Data Not Available (CI, PHP, MySQL)

    - by tegaralaga
    First of all, i do apologize for my bad english, cos english isn't my native language. I want to display yearly report based on month, let say i got order on January, August, December, but the rest there's no order. So in MySQL database only have 3 order (Jan,Aug,Dec). When i query use CI ( select month(order_date) as month_name , count(order_id) as amount from order where year(order_date)=2011 group by month(order_date) ) there's only 3 data let say the 3 data is (use $query-result_array()) Array ( [0] => Array ( [month_num] => 1 [amount] => 4 ) [1] => Array ( [month_num] => 8 [amount] => 1 ) [2] => Array ( [month_num] => 12 [amount] => 19 ) ) how to make it to 12 data (12 Month) the array become like this (when data not available the amount is 0) Array ( [0] => Array ( [month_num] => 1 [amount] => 4 ) [1] => Array ( [month_num] => 2 [amount] => 0 ) [2] => Array ( [month_num] => 3 [amount] => 0 ) etc ) Thanks in advance :)

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  • (PHP) Converting an array of arrays from one format into another

    - by Richard Carter
    Hi, I currently have an array, created from a database, an example of which looks like the following: Array( [0] => Array ( objectid => 2, name => title, value => apple ), [1] => Array ( objectid => 2, name => colour, value => red ), [2] => Array ( objectid => 3, name => title, value => pear ), [3] => Array ( objectid => 3, name => colour, value => green ) ) What I would like to do is group all the items in the array by their objectid, and convert the 'name' values into keys and 'value' values into values of an associative array....like below: Array ( [0] => Array ( objectid => 2, title => apple, colour => red ), [1] => Array ( objectid => 3, title => pear, colour => green ) ) I've tried a few things but haven't really got anywhere.. Any ideas? Thanks in advance

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  • Errors checking and opening a URL using PHP

    - by Jean
    Hello Here is one script with out any errors $url="http://yahoo.com"; $file1 = fopen($url, "r"); $content = file_get_contents($url); $t_beg = explode('<title>',$content); $t_end = explode('</title>',$t_beg[1]); echo $t_end[0]; And here is the same script using a look to check multiple urls and getting errors for ($j=1;$j<=$i;$j++) { if ($x[$j]!=''){ $t_u = "http:".$x[$j]; $file2 = fopen($t_u, "r"); $content2 = file_get_contents($t_u); $t_beg = explode('<title>',$content); $t_end = explode('</title>',$t_beg[1]); echo $t_end[0]; } } The error is Warning: fopen() [function.fopen]: php_network_getaddresses: getaddrinfo failed: No such host is known. in g:/ What exactly is wrong here?

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  • Searching with MATCH(), AGAINST() and AS score with mysqli and php

    - by Drew
    Below is the code I am using to search my table. I have made the relevant columns FULLTEXT in the table. This doesn't return me anything. Can someone tell me what it is that i'm doing wrong? Thanks in advance. $sql = 'SELECT id,uname,class,school, MATCH(uname, class, school) AGAINST(?) AS score FROM images WHERE MATCH(uname, class, school) AGAINST(? IN BOOLEAN MODE) ORDER BY score DES'; $stmt = $db_connection->prepare($sql); $stmt->bind_param('ss',$keyword,$keyword); $stmt->execute(); $stmt->store_result(); $stmt->bind_result($id,$uname,$class,$school); $xml = "<data>".PHP_EOL; while($stmt->fetch()){ $xml .= " <person>".PHP_EOL; $xml .= " <id>$id</id>".PHP_EOL; $xml .= " <name>$uname</name>".PHP_EOL; $xml .= " <class>$class</class>".PHP_EOL; $xml .= " <school>$school</school>".PHP_EOL; $xml .= " </person>".PHP_EOL; } $xml .= "</data>"; echo $xml; Below is an image of the indexes of the table:

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  • PHP & MySQL Image deletion problem?

    - by IMAGE
    I have this script that deletes a users avatar image that is stored in a folder name thumbs and another named images and the image name is stored in a mysql database. But for some reason the script deletes all the users info for example if the users id is 3 all of the users info like first name last name age and so are deleted as well, basically everything is deleted including the user how do I fix this so only the images and image name is deleted? Here is the code. $user_id = '3'; if (isset($_POST['delete_image'])) { $a = "SELECT * FROM users WHERE avatar = '". $avatar ."' AND user_id = '". $user_id ."'"; $r = mysqli_query ($mysqli, $a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($mysqli)); if ($r == TRUE) { unlink("../members/" . $user_id . "/images/" . $avatar); unlink("../members/" . $user_id . "/images/thumbs/" . $avatar); $a = "DELETE FROM users WHERE avatar = '". $avatar ."' AND user_id = '". $user_id ."'"; $r = mysqli_query ($mysqli, $a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($mysqli)); } }

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  • PHP + MySQL - Match first letter of directory

    - by user1822825
    Let's say I have a class table. In the class table, there are many students with their pictures. In the first registration, I've registered the class and students with pictures. The pictures were put into a directory like classid_classname. Then, I change the class name. Now, I'm adding the student's picture. Now, the new picture can't be recognized because the class name has changed. The pic url will be set as classid_class(new)name. How can I match the first letter of the directory? This is my update code : $classID= $_POST["classID"]; $className= $_POST["className"]; $p1 = $_FILES['p1']['name']; $p2 = $_FILES['p2']['name']; $p3 = $_FILES['p3']['name']; $direct = $_POST["className"]; $direct = strtolower($direct); $direct = str_replace(' ', '_', $direct); $tfish = $classID."_".$direct; //the directory variable will have new name because it can't be fetched if the directory has been changed many times// $file = "slider_imagesClass/".$tfish."/"; $url = "/".$tfish."/"; How can I make the variable to match the first letter of the directory because the classID will not change? Thank you. Really appreciate your help :D

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  • SSI or PHP Include()?

    - by Ozzy
    Hi all, basically i am launching a site soon and i predict ALOT of traffic. For scenarios sake, lets say i will have 1m uniques a day. The data will be static but i need to have includes aswell I will only include a html page inside another html page, nothing dynamic (i have my reasons that i wont disclose to keep this simple) My question is, performance wise what is faster or

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  • How to merge objects in php ?

    - by The Devil
    Hey everybody, I'm currently re-writing a class which handles xml files. Depending on the xml file and it's structure I sometimes need to merge objects. Lets say once I have this: <page name="a title"/> And another time I have this: <page name="a title"> <permission>administrator</permission> </page> Before, I needed only the attributes from the "page" element. That's why a lot of my code expects an object containing only the attributes ($loadedXml-attributes()). Now there are xml files in which the <permission> element is required. I did manage to merge the objects (though not as I wanted) but I can't get to access one of them (most probably it's something I'm missing). To merge my objects I used this code: (object) array_merge( (array) $loadedXml->attributes(), (array) $loadedXml->children() ); This is what I get from print_r(): stdClass Object ( [@attributes] => Array ( [name] => a title ) [permission] => Array ( [0] => administrator ) ) So now my question is how to access the @attributes method ? Thanks in advance, The Devil

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  • <?php echo $questionTitle; ?> [closed]

    - by oil
    So this "question" makes no sense and will be closed. But are you willing to spend your hard-earned reputation to downvote it? Go on. A programmer should be able to make logical decisions, why will you downvote the question then instead of ignoring it?

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  • Math - Adding with PHP

    - by Wayne
    Basically I can't get it right. I need something like this: if($p == 1) { $start = 0; $limit = 16; } The numbers must add on depending on the value of the $p, e.g. if $p is 5 then the values of $start and $limit would be: if($p == 5) { $start = 64; $limit = 80; } The math is to add 16, depending on the value of $p. Thanks.

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  • count and fetch rows in php

    - by Mac Taylor
    hey guys i have a table in my mysql database named (names) now everyone can save their real names now i want to query this table and find out how many times these names used forexample the output should be : Jakob (20) Jenny (17) now this is my own code : list($usernames) =mysql_fetch_row(mysql_query('SELECT name FROM table_user GROUP BY name ORDER BY COUNT(name) DESC LIMIT 50 ')); list($c) =mysql_num_rows(mysql_query('SELECT COUNT(name) FROM table_user GROUP BY name ')); print $usernames.'('.$c.')' is this a correct approach ?!

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  • php extract data from mysql

    - by florin
    I have this mysql table with the following rows: id_cont suma_lun month year -------------------------------------------- FL28 2133 March 2012 FL28 2144 April 2012 FL28 2155 May 2012 FL28 2166 June 2012 How can i extract suma_lun, month and year foreach id_cont? so that i get an output like this: ID: Month: Monthly Sum: Year: ---------------------------------------------- FL28 March 2133 2012 April 2144 2012 May 2155 2012 June 2166 2012 This is my current code: $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) die ('Could not connect to database: '.mysql_error()); mysql_select_db(DB_DATABASE,$link); $sql="SELECT * FROM test WHERE id_cont = '$cur'"; $result=mysql_query($sql); while ($row=mysql_fetch_array($result)) { $a=$row["id_cont"]; $b=$row["suma_lun"]; $c=$row["month"]; $d=$row["year"]; } I echo the data in a table Thanks!

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  • PHP: saving multiple tuples to the same table

    - by Binaryrespawn
    Hi all, I am trying to save data into the same table several times from the same $_POST array. You see, on my registration form a family is allowed to specify the names, gender and age of all members they wish to bring along. This data is being stored in a database table called aditional_member. Based on family size, I am dynamically generating html table rows where each row has the number of input fields required. <tr id="additionalmember_3"> <td><input type="text" name="first_name1" maxlength="50" value="" /></td> <td><input type="text" name="last_name1" maxlength="50" value="" /></td> td><select name="gender1" value="" ><option value='Male'>Male</option><option value='Female'>Female</option></select></td> <td><select name="age_group"value="" ><option value='18 to 30'>18 to 30</option><option value='31 to 60'>31 to 60</option></select></td> </tr> When I POST, let us say I have three sets of input(3 table rows as above), Is there a simple way to insert into the addional_member table the data from row one, then row two, and so on. I tried looping the insert statement, which failed. I am basically open to suggestions, I do not need to use a table. Any ideas, Thank you.

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  • PHP weirdness extending IMagick class

    - by Jamie Carl
    This is a really weird one. I have some code that is happily working on version 2.1.1RC1 of the php5-imagick module. It's basically just a class I wrote that extends the Imagick class and manages images stored in a database. Since upgrading to version 3.0.0RC1 (thankfully only on my dev box) things have gone to hell. It seems that object members are writeable but are NOT readable. Take the following sample code: class db_image extends IMagick { private $data; function __construct( $id = null ){ parent::__construct(); $this->data = 'some plain text'; echo $this->data; } This will output absolutely NOTHING. My debugger indicates that the contents of $this-data are the correct string value, but I am unable to read the value back out of the member variable. Seriously. WTF? Does anyone know what is causing this or has seen it before? I don't even know how to replicate this behaviour in my own classes.

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  • Capitalize Words in PHP with custom delimitor

    - by Paulie
    Hey guys, i need a method to capitalize every first letter of a word. This is what i got so far and it is working for almost every string...but it fails on this one "WELLNESS & RENOMME". // method in stringModify Class function capitalizeWords($words, $charList) { $capitalizeNext = true; for ($i = 0, $max = strlen($words); $i < $max; $i++) { if (strpos($charList, $words[$i]) !== false) { $`capitalizeNext` = true; } else if ($capitalizeNext) { $capitalizeNext = false; $words[$i] = strtoupper($words[$i]); } } return $words; } // Calling method $stringModify->capitalizeWords("WELLNESS & RENOMME", " -&"); I hope someone can help me out...i tried for 1,5 hours now and don't have a clue. Thanks in advance for any tips or hints. Greetz Paulie

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  • change last key name from array in php

    - by robertdd
    i want to be able to change the last key from array i try with this function i made: function getlastimage($newkey){ $arr = $_SESSION['files']; $oldkey = array_pop(array_keys($arr)); $arr[$newkey] = $arr[$oldkey]; unset($arr[$oldkey]); $results = end($arr); //echo json_encode($results); print_r($arr); } if i call the function getlastimage('newkey') it change the key!but after if i print the $_SESSION the key is not changed? why this?

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  • PHP form processing - how to capture text from field that has variable Name/ID

    - by user80151
    I have a form that has a field pulled from the database as a dropdown. I need to get the text selected in the dropdown but I don't know in advance what the field ID will be. This is basically just a form that has already been generated. I don't need to pull anything from the database, it's already on this page. All I need to do is get the form information and email it, no writing to the database. I know how to do the _Request for the other fields based on the ID but I'm not sure how to do this one. The ID changes. It can be ID=1, ID-2, etc. I need to do something like: _REQUEST form element where ID is LIKE "ID[*]" or something similar. Any suggestions or links to tutorials? Here are a couple samples of what the dropdown renders on the page: <div class="wrapperAttribsOptions"> <h4 class="optionName back"><label class="attribsSelect" for="attrib- 1">Model</label></h4> <div class="back"> <select name="id[1]" id="attrib-1"> <option value="45">VC3-4C</option> <option value="1">VC3-4PG</option> <option value="3">VC3-4SG</option> <div class="wrapperAttribsOptions"> <h4 class="optionName back"><label class="attribsSelect" for="attrib-14">SPK Model</label></h4> <div class="back"> <select name="id[14]" id="attrib-14"> <option value="43">SPK-4</option> <option value="44">SPK-8</option> </select> TIA

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  • PHP/MYSQL Trouble Selecting by Primary Key

    - by djs22
    Hi all, So I have a primary key column called key. I'm trying to select the row with key = 1 via this code: $query ="SELECT * FROM Bowlers WHERE key = '1'"; $result = mysql_query($query) or die(mysql_error()); For some reason, I'm getting this result: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'key = '1'' at line 1 The mysql statement works for using other keys, ie WHERE name = 'djs22'. Any ideas?

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