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PHP - Error: The Argument Should Be An Array
I'm getting this error, but I'm not sure how to fix it (on my Wordpress site).
Warning: usort() [function.usort]: The argument should be an array in /home2/lushwebs/public_html/4mulaevents/wp-content/themes/dailyedition/includes/talking-points.php on line 20 Warning: Invalid argument supplied for foreach() in /home2/lushwebs/public_html/4mulaevents/wp-content/themes/dailyedition/includes/talking-points.php on line 22 Any help appreciated. Thanks! Similar TutorialsI am having some problems getting a query correct. Basically I have two tables, one with listings and another with categories. In the listings table I have a column called shortdescription. I am trying to pull the shortdescription from the listings table with the query $shortdesc = array_shift(mysql_fetch_row(mysql_query("select shortdescription from links where category = '".$scat->id."' "))); The shortdecription display properly on the pages display the listings, however I am getting the following error on any pages that only display the parent categories Warning: array_shift() [function.array-shift]: The argument should be an array in /home/...path/file.php on line 1462 The listings id numbers begin at 75+ because the initial parent category id ends at 74. The query seems to be searching for listing ids below 75 and spitting out an error because it is not finding them. Any ideas on how to eliminate this error and/or stop the query from looking for non-existant data? For some reason I keep getting a recurring error every time someone uses a page on my site. Here is the error: Code: [Select] [Date] PHP Warning: array_key_exists() [<a href='function.array-key-exists'>function.array-key-exists</a>]: The second argument should be either an array or an object in /home/page.php on line 52 The offending line is he Code: [Select] echo array_key_exists($i, $alpha) ? '<li><a href="?a=show&letter='.$i.'&type='.$type.'">'.$i.'</a></li> ' : '<li class="no_highlight">'.$i.'</li> The code works, but the error message must be showing that I am performing the echo incorrectly. Anyone have any ideas on where I could be going wrong (other than the exact line, that is )? Here's the rest of the code for full reference (the offending line is near the top). Thanks for any help! Code: [Select] <?php $sql = "SELECT *, UPPER(SUBSTRING(series,1,1)) AS letter FROM design WHERE type = '$type' ORDER BY series"; $query = mysql_query($sql) or die(mysql_error()); // Display a range of letters. echo '<ul class="alphabet">'; while ($records = mysql_fetch_assoc($query)) { $alpha[$records['letter']] += 1; ${$records['letter']}[$records['series']] = $records['series']; } foreach(range('A','Z') AS $i) { echo array_key_exists($i, $alpha) ? '<li><a href="?a=show&letter='.$i.'&type='.$type.'">'.$i.'</a></li> ' : '<li class="no_highlight">'.$i.'</li> '; echo $i != 'Z' ? ' ':'</ul>'; } // Display records of the letter selected. if (isset($_GET['letter']) && !empty($_GET['letter']) && !is_numeric($_GET['letter'])) { $letter = substr($_GET['letter'], 0, 1); } else { $letter = ''; } $a = (isset($_GET['a'])) ? strip_tags(trim(htmlentities($_GET['a'], ENT_QUOTES))) : 'index'; switch ($a) { case 'index': $select_series = mysql_query("SELECT *, UPPER(SUBSTRING(series,1,1)) AS letter FROM design WHERE `type` = '$type' ORDER BY `id` DESC LIMIT 8") or die (mysql_error()); $i=3; echo '<h3>Newest Additions</h3>'; while($row2 = mysql_fetch_array($select_series)) { extract ($row2); $contents_here = '<div class="envelope"><a href="?a=show&letter='.$letter.'&type='.$type.'#'.$title.'"><img src="previews/'.$id.'.jpg" class="tips" title="'.$title.'|'.$series.'"><br>'.$title.'</a></div>'; if ($i==0) { echo ''.$contents_here.''; } else if ($i==3) { echo ''.$contents_here.''; } else { echo ''.$contents_here.''; } $i++; $i=$i%3; } break; case 'show': $letter2 = (isset($_GET['letter'])) ? strip_tags(trim(htmlentities($_GET['letter'], ENT_QUOTES))) : ''; $sql = "SELECT * ,UPPER(SUBSTRING(series,1,1)) AS letter, date_format(design.datetime, '%M %d, %Y') AS datetime FROM design WHERE series LIKE '$letter2%' AND type = '$type' ORDER BY series ASC, id ASC"; $query = mysql_query($sql) or die (mysql_error()); echo '<h3><a name="'.$i.'"><b>'.$letter.'</b></a></h3> <table align="center" width="100%"><tr><td>'; $series = ''; while ($row = mysql_fetch_assoc($query)) { $image = $row[image]; $datetime = $row[datetime]; $id = $row[id]; $type = $row[type]; $title = $row[title]; $count_display = $row[count_display]; if ($series != $row['series']) { //Assign category to holding var $series = $row['series']; //echo Category heading echo '<h3><a href="#" class="top"></a>'.$series.'</h3> '; } echo '<table align="center" class="format" cellpadding="0" cellspacing="0"> <tr><td> <img src="previews/'.$id.'.jpg" align="right"> <td class="design"><ul class="design_info"> <li class="special"><b>Title:</b> <a name="'.$title.'">'.$title.'</a> <ul> <li><b>Series:</b> '.$series.' <li><b>Type:</b> '.$type.' <li><b>Uploaded:</b> '.$datetime.' <li><b>Downloads:</b> '.$count_display.' </ul></ul> <ul class="design_dl"> <li class="preview"><a href="graphics/preview.php?id='.$id.'" target="_blank">preview</a> <li class="download"><a href="graphics/download.php?id='.$id.'">download</a> </ul> </td></tr> </table>'; } } echo '</table>'; ?> Hi, I am attempting to export some table data to an excel spreadsheet. I am having trouble with the final few lines and getting the following error within the spreadsheet (which is otherwise working well). Quote The argument should be an array in/home/website/public_html/student_list_export.php on line 34 I get the same error message for the line below also (echo implode("\t", array_values($row)) . "\n". I'm at a loss. This seems like a pretty obvious and direct error message but I can't get it working properly. Can somebody please help me? Below is the code I am working with. <?php include('includes/admin_session.php'); require_once("includes/connection.php"); function cleanData(&$str) { $str = preg_replace("/\t/", "\\t", $str); $str = preg_replace("/\r?\n/", "\\n", $str); if(strstr($str, '"')) $str = '"' . str_replace('"', '""', $str) . '"'; } // file name for download $filename = "student_list_" . date('Ymd') . ".xls"; header("Content-Disposition: attachment; filename=\"$filename\""); header("Content-Type: application/vnd.ms-excel"); $flag = false; $result = mysqli_query($connection,"SELECT * FROM students ORDER BY tutor") or die(mysqli_error($connection)); while(false !== ($row = mysqli_fetch_assoc($result))) { if(!$flag) { // display field/column names as first row echo implode("\t", array_keys($row)) . "\n"; $flag = true; } array_walk($row, 'cleanData'); echo implode("\t", array_values($row)) . "\n"; } ?> Im getting a warning with in_array.I have used if() inside for loop to check if value has already been printed or not. If not it will print and save the value in $done[].Can someone please point me out my mistake that i am getting a warning. Code: [Select] <?php for($i=0;$i<=$count;$i++) { $count_final = explode("_", $result[$i]); if (in_array($count_final[0],$done)) { //Line 480 echo "<tr>"; echo"<td>$count_final[0]</td>"; $templ = $count_final[0]."_left"; $tempr = $count_final[0]."_right"; $cleft = $count_middle[$templ]; $cright = $count_middle[$tempr]; echo"<td>$cright</td>"; echo"<td>$cleft</td>"; echo "</tr>"; $done[] = $count_final[0]; } } ?> Code: [Select] Warning: in_array() [function.in-array]: Wrong datatype for second argument in C:\Users\\Desktop\yyy.php on line 480 This is the full error...Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in This is what I am trying to do... 1. get data from the page before which it gets fine in the $_GET function... 2. then set the sersion[username] to $username...(I think I am doing something wrong there) 3. then if the $username doesnt match the username pulled from the database (according to the $_GET info) then open a different databse and update a value in that table. 4. then kill the code die (); someone please help I have been trying to figure this out all day Code: [Select] <?php $username = $_SESSION['username']; $deleted = $_GET['deleted']; // Connect to MySQL $connect = mysql_connect("db","user","pass") or die("Not connected"); mysql_select_db("dv") or die("could not log in"); // grab the information according to the isbn number $query = "SELECT * FROM 'boox' WHERE isbn='$deleted'"; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { if ($username==$row['username']) { echo"Welcome, ".$_SESSION['username'].""; // Delete entry where isbn number matches accordingly mysql_query("DELETE * FROM 'boox' WHERE isbn='$deleted'"); } elseif ($username!=$row['username']) { $username = $_SESSION['username']; $connect = mysql_connect("db","user","pass") or die("Not connected"); mysql_select_db("db") or die("could not log in"); // run a query to deactivete a account $acti = mysql_query("UPDATE 'desiredusers' SET activated='2' WHERE username='$username'"); $byebye = "SELECT * FROM 'desiredusers' WHERE isbn='$deleted'"; $results = mysql_query($byebye); while($row2 = mysql_fetch_array($results)) echo "here".$username."here"; die("Your account is deactivated contact College Boox Store to get your account back."); } else echo "there is nothing being checked" . $username . ""; } ?> I have this search script, that finds a username in a table and displays it. <form name="search" method="post" action="<?=$PHP_SELF?>"> Search: <input type="text" name="username" /> <input type="hidden" name="searching" value="yes" /> <input type="submit" name="search" value="Search" /> </form> <? //This is only displayed if they have submitted the form $searching = $_POST['searching']; $find = $_POST['username']; if ($searching =="yes") { echo "<h2>Results</h2><p>"; //If they did not enter a search term we give them an error if ($find == "") { echo "<p>You forgot to enter a search term"; exit; } // Otherwise we connect to our Database // We preform a bit of filtering $find = strtoupper($find); $find = strip_tags($find); $find = trim ($find); //Now we search for our search term, in the field the user specified $data = mysql_query("SELECT username FROM users WHERE upper($find) LIKE'%$find%'"); //And we display the results while($result = mysql_fetch_array( $data )) { echo $result['username']; echo "<br>"; echo "<br>"; } //This counts the number or results - and if there wasn't any it gives them a little message explaining that $anymatches= mysql_num_rows ($data); if ($anymatches == 0) { echo "Sorry, but we can not find an entry to match your query<br><br>"; } //And we remind them what they searched for echo "<b>Searched For:</b> " .$find; } ?> For some reason it does not work and gives this message: [b]Warning[/b]: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in [b]C:\Program Files\EasyPHP5.2.10\www\yanille\users.php[/b] on line [b]153[/b] Error: Unknown column 'MI' in 'where clause' The reason it says MI is because I searched mi in the database which is part of the username in mike123 What would be the problem? I think I did make the query correct. Didn't I? Thanks for the help ahead of time. Hello I'm using tcpdf to convert some stuff to pdf. On my local computer running wamp my script works perfect and the pdf is generated without problems When i put my files in a webserver i got this error: Warning: in_array() [function.in-array]: Wrong datatype for second argument in /home/vieira/public_html/topdf/dbcon.class.php on line 68 Line 68 is: if(in_array($key,$arr_switches)){ //check switches My complete code is: // Get fields of enum on/off switches type $result2 = mysql_query("SHOW FIELDS FROM $db_name.$table"); $counter=0; while($row = @mysql_fetch_array($result2)){ //echo $row['Field'] . ' ' . $row['Type']."<br/>"; if($row['Type'] === "enum('on','off')"){ $arr_switches[$counter] = $row['Field']; $counter++; } } //print"<pre>"; print_r($arr_switches); print"</pre>";exit; $counter = 0; while($row = @mysql_fetch_array($result)){ foreach( $row as $key => $val ){ if(!is_numeric($key)) { $row_rs_certidao[$key] = $val; if(in_array($key,$arr_switches)){ //check switches $record_key[$counter] = htmlentities('<?php if (!(strcmp($row_rs_certidao['."'".$key."'".'],"on"))) {echo "x";} ?>'); if($val==='on') $record_val[$counter] = "x"; //turn on switches else $record_val[$counter] = ''; //turn off }else{ $record_key[$counter] = htmlentities('<?php echo $row_rs_certidao['."'".$key."'".']; ?>'); $record_val[$counter] = htmlentities($val); } $counter++; } } } Anyone can help?
<!DOCTYPE html>
{ hi there, i am fairly new to OOPs in php, i get an error when i declare the argument type (as object) in a function and pass the same type (object). class eBlast { public static function getEmail(object $result) { return $result->email; } } $r = mysql_fetch_object($query); eBlast::getEmail($r); echo gettype($r); // outputs: object error is : Code: [Select] Catchable fatal error: Argument 1 passed to eBlast::getEmail() must be an instance of object, instance of stdClass given, called in C:\wamp\www\integra\client\pl_eblast\admin\send_emails.php on line 145 and defined in C:\wamp\www\integra\client\pl_eblast\app\app.eBlast.php on line 8 if i remove the type declaration in the function it works, but just would like to know why it shows error when pass the same type, also isnt mysql_fetch_object is the instance of stdclass? thanks in advance! I'm getting an invalid argument error for my While loop. I can't figure out why. Code: [Select] $sql="SELECT * FROM schools ORDER BY class DESC,sectional ASC,team ASC"; $results = mysql_query($sql); while ($team = mysql_fetch_assoc($results)) { echo $team['school'] . '<br>'; } Here is the error: Quote Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/******/public_html/hoosierhoopsmagazine.com/resources/sectional_assignments.php on line 7 I have a table with a list of users and an edit button and delete button. When the edit button is pressed on a site it passes the user_id as p_id to the page that catches it and displays the data. The problem is when I click on the "update user" button, I get the following error:
Warning: Undefined variable $the_user_id in C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php on line 10 The weird thing is I had another update user page with a table I created that ran the query to update the table in the database just fine. But as I created it, it didn't look all that great so I recreated the page and used a bootstrap table because of the much cleaner look. Both pages have the exact same PHP code, the only difference is the bootstrap table I added in. So I'm really at a loss with this. Other than the table and PHP code, there is a script at the bottom of the page for the table itself to allow for searching within the table, i'll include that as well. The PHP code is as follows:
<?php //THE "p_id" IS BROUGHT OVER FROM THE EDIT BUTTON ON VIEW_ALL_USERS if (isset($_GET['p_id'])) { $the_user_id = $_GET['p_id']; } // QUERY TO PULL THE SITE INFORMATION FROM THE p_id THAT WAS PULLED OVER $query = "SELECT * FROM users WHERE user_id = $the_user_id "; $select_user = mysqli_query($connection,$query); //SET VALUES FROM ARRAY TO VARIABLES while($row = mysqli_fetch_assoc($select_user)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_email = $row['user_email']; $user_phone = $row['user_phone']; //$user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id']; } THE UPDATE QUERY CODE....................................................................................................................
<?php if(isset($_POST['update_user'])) { $user_id = $_POST['user_id']; $user_firstname = $_POST['user_firstname']; $user_lastname = $_POST['user_lastname']; $username = $_POST['username']; $user_email = $_POST['user_email']; $user_phone = $_POST['user_phone']; //$user_image = $_POST['user_image']; $user_title_id = $_POST['user_title_id']; $user_role_id = $_POST['user_role_id'];
$query = "UPDATE users SET "; $query .= "user_id = '{$user_id}', "; $query .= "user_firstname = '{$user_firstname}', "; $query .= "user_lastname = '{$user_lastname}', "; $query .= "username = '{$username}', "; $query .= "user_email = '{$user_email}', "; $query .= "user_phone = '{$user_phone}', "; //$query .= "user_image = '{$user_image}', "; $query .= "user_title_id = '{$user_title_id}', "; $query .= "user_role_id = '{$user_role_id}' "; $query .= "WHERE user_id = '{$the_user_id}' "; $update_user = mysqli_query($connection,$query); if(! $update_user) { die("QUERY FAILED" . mysqli_error($connection)); } } ?> THE "UPDATE USER" BUTTON THE USER CLICKS ON TO UPDATE....................................................................................................................
<div class="col-1"> <button class="btn btn-primary" type="submit" name="update_user">Update User</button> </div>
Any Help is Greatly Appreciated! Edited March 23 by ZsereneMy script IS working, but I can't get around a blank array error when no errors exist. Below is my code, and as you can see, I am being handed this: Notice: Undefined variable: error in C:\wamp\www\php\form_validation.php on line 19 as a result of my ... if (is_array($error)) { ... .. I could do if (@is_array($error)) { (note the @), but I hate using that thing... I've tried several things with no luck, so any ideas welcome at this point. <?php if (isset($_POST['set_test'])) { if (!preg_match("/^[A-Za-z' -]{1,50}$/", $_POST['first_name'])) { $error[] = "Please enter a valid First Name"; } if (!preg_match("/^[A-Za-z' -]{1,50}$/", $_POST['last_name'])) { $error[] = "Please enter a valid Last Name"; } if (is_array($error)) { foreach ($error as $err_message) { echo $err_message . "<br />"; } } } ?> i keep getting this error when there is no array. Warning: array_values() [function.array-values]: The argument should be an array in /home/learnat2/public_html/cart.php on line 37 but i put this code to prevent the error Code: [Select] $items = $_SESSION['items']; if(is_array($items)) { $quantity = array_count_values($items); $items = array_unique($items); } i still get the error. it should not display an error because $items is not an array, no values have been added. SOLVED Hi guys this one page is giving me a headache. i posted earlier today with a question on how to insert multiple records and got given a quick demo of how to do it so ive gone away to look more and ive edited my code. however it still doesnt work. im wondering if ive coded a tiny bit wrong. id be really appreciative if you guys could help me out Code: [Select] <?php include 'dbc.php'; page_protect(); company(); $stafflist = mysql_query("SELECT * FROM StaffList WHERE full_name != 'Adam Carter' AND full_name != 'Jakata' AND branch = '$_SESSION[branch]' "); $id = $_POST['id']; if (isset($_POST['submit'])) { //Assign each array to a variable $id = $_POST['id']; $user_name = $_POST['user_name']; $Serviceamount = $Serviceamount['item']; $limit = count($id); echo $limit; $values = array(); // initialize an empty array to hold the values for($k=0;$k<$limit;$k++){ $user_name[$k] = check_input($user_name[$k]); $Serviceamount[$k] = check_input($Serviceamount[$k]); $values[$k] = "( '{$id[$k]}', '{$Serviceamount[$k]}')"; // build the array of values for the query string } $query = "INSERT INTO `Services` (Staffname, ServiceAmount) VALUES " . implode( ', ', $values ); // Form the query string and add the implod()ed values if (!mysql_query($query,$link)){ die('Error: ' . mysql_error()); } else { echo "$row record added"; } echo $values; } if (checkAdmin()) { ?> <html> <head> <title>Book Off Holiday</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <script src="php_calendar/scripts.js" type="text/javascript"></script> <link href="styles.css" rel="stylesheet" type="text/css"> </head> <body> <form name="form" action="Newkpi.php" method="post"> <table width="100%" border="0" cellspacing="0" cellpadding="5" class="main"> <tr> <td colspan="3"> </td> </tr> <td width="160" valign="top"> <?php if (isset($_SESSION['user_id'])) { } ?> <a href="admin.php">Admin CP </a> </td> <td width="732" valign="top"> <p> <h3 class="titlehdr">New KPI</h3> <table width="300px" border="0" align="Centre" cellpadding="2" cellspacing="0"> <tr bgcolor="#000050"> <td width="20px"><h3 class="Text2">Staff ID</h3></td> <td width="20px"><h3 class="Text2">Staff Member</h3></td> <td width="20px"><h3 class="Text2">Service Amount</h3></td> <td width="20px"><h3 class="Text2">Service Date</h3></td> <td width="20px"><h3 class="Text2">Forecast For Next Month</h3></td> <td width="20px"><h3 class="Text2">Product Sales</h3></td> <td width="20px"><h3 class="Text2">Clients This Month</h3></td> <td width="20px"><h3 class="Text2">Personel Retension</h3></td> <td width="20px"><h3 class="Text2">Total Retension</h3></td> <td width="20px"><h3 class="Text2">Colours</h3></td> <td width="20px"><h3 class="Text2">Cuts</h3></td> <td width="20px"><h3 class="Text2">Pre-Booking</h3></td> <td width="20px"><h3 class="Text2">Time Used</h3></td> </tr> <?php while ($rrows = mysql_fetch_array($stafflist)) {?> <tr> <td><h3 class="Text3"><input type="" name="id[]" id="id[]" size="4" value="<?php echo $rrows['id'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="user_name[]" id="user_name[]" value="<?php echo $rrows['full_name'];?>" /></h3></td> <td><h3 class="Text3"><input name="Serviceamount" type="text" size="4" id="Serviceamount"></h3></td> <td><h3 class="Text3"><input name="servicedate" type="text" size="4" id="servicedate"></h3></td> <td><h3 class="Text3"><input name="forecast" type="text" size="4" id="forecast"></h3></td> <td><h3 class="Text3"><input name="productsales" type="text" size="4" id="productsales"></h3></td> <td><h3 class="Text3"><input name="Clientsthismonth" type="text" size="4" id="Clientsthismonth"></h3></td> <td><h3 class="Text3"><input name="Personelret" type="text" size="4" id="Personelret"></h3></td> <td><h3 class="Text3"><input name="Totalret" type="text" size="4" id="Totalret"></h3></td> <td><h3 class="Text3"><input name="colours" type="text" size="4" id="colours"></h3></td> <td><h3 class="Text3"><input name="cuts" type="text" size="4" id="cuts"></h3></td> <td><h3 class="Text3"><input name="prebooking" type="text" size="4" id="prebooking"></h3></td> <td><h3 class="Text3"><input name="timeused" type="text" size="4" id="timeused"></h3></td> </tr> <?php } ?> </table> <input name="submit" type="submit" id="submit" value="Create"> </table> </form> </body> </html> <?php } ?> thank you in advance I've been thinking of using an assoc_array for validation code in php to have an associative of all the fields (name, email, message) and to see if there has been an "ERROR" declared in any of them? How would I first declare this array? and How would I check to see if they are empty or any of them contain "ERROR". Thanks Im getting the following error on this line: Code: [Select] Fatal error: Cannot use string offset as an array Code: [Select] $array['headers_style'] = 'color:'.$color['headers_text']['text'].'; background-color:'.$color['headers']['background'].';'; This error only occurs on one page, others that use this line dont have a problem. So i tried to delete parts of the code from the page causing the error but it still exists. So i deleted a huge peice of the code and the error vanished. So i eliminated portions of the huge code 1 at a time but the error stays until i delete all of the code. Anyone know why this page doesnt like the error? i could post the code but its a huge chunk(around 600 lines) Hi guys, I am pretty new to php and mysql, so please don't only tell me what to do without showing me. Would be very appreciated. Thanks in advance!! Here is what I am trying to do. I am trying to use php to print the fields a table in Mysql into a form page called index.php. I think I sucessfully did it, but not the best way (mixing php and html, because I couldn't figure it out how to do it seperately) Everything looks good, except when I press submit and now we go to the insert_stats.php where the error: Warning: Invalid argument supplied for foreach() in /insert_stats.php on line 8 insert_stats.php Code: [Select] <? include("sql.php"); $paddler = $_POST['paddler'] ; $practice = $_POST['practice'] ; $header = $_POST['header'] ; [b]foreach ($header as $value) [/b] { $insert="INSERT INTO pushup ($header) VALUES ('$value')"; mysql_query($insert) OR die(mysql_error()) ; } $insert2="INSERT INTO pushup (paddler_id , practice_id) VALUES ('$paddler' , '$practice')"; mysql_query($insert2) OR die(mysql_error()) ; mysql_close(); ?> index.php Code: [Select] <table width="250" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form id="pushup" name="pushup" method="post" action="insert_stats.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td colspan="3" bgcolor="#E6E6E6"><strong>IDENTIFICATION</strong> </td> </tr> <tr> <td width="80%">Paddler ID</td> <td width="2%">:</td> <td width="18%"><input name="paddler" type="text" id="paddler" size="1" /></td> </tr> <tr> <td>Practice #</td> <td>:</td> <td><input name="practice" type="text" id="practice" size="1" /></td> </tr> <tr> <td colspan="3" bgcolor="#E6E6E6"><strong>ADD FITNESS STATS</strong> </td> </tr> <? include("sql.php"); $result = mysql_query("SELECT * FROM pushup"); $printed_headers = false; while ( $row = mysql_fetch_array($result) ) { if ( !$printed_headers ) { //print the headers once: echo "<tr>"; foreach ( array_keys($row) AS $header ) { if($header == 'paddler_id' || $header == 'practice_id' || $header == 'p_id') continue; if ( !is_int($header) ) { echo "<td>$header</td><td>:</td><td><input name='".$header."' type=\"text\" id='".$header."' size=\"1\" /></td></tr>"; } } $printed_headers = true; } } mysql_free_result($result); mysql_close(); ?> <tr> <td><input type="reset" name="Submit2" value="Reset" /></td> <td> </td> <td><input type="submit" name="Submit" value="Submit" /></td> </table> </td> </form> </tr> </table> database.php
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$res = pg_query ($conn, "SELECT artist, composer, genre, title, album, label, price, description FROM music");
echo "<table border='1'>"; while($a = pg_fetch_array($res)){ echo "<td>" . $a['1'] . "</td>"; echo "<td>" . $a['2'] . "</td>"; echo "<td>" . $a['3'] . "</td>"; echo "<td>" . $a['4'] . "</td>"; echo "<td>" . $a['5'] . "</td>"; echo "<td>" . $a['6'] . "</td>"; echo "<td>" . $a['7'] . "</td>"; echo '<td><input type="checkbox" name="music[]" value="$ref"' . $row['0'] . $row['1'] . $row['2'] . $row['3'] . $row['4'] . $row['5'] . $row['6'] . $row['7'] . '"/></td>'; echo"</tr>"; } echo "</table>\n"; ?> _____________________________________________________________________________- shoppingbasket.php _____________________________________________________________________________- <form action="shoppingbasket.php" method="POST"> Send To Basket: <name="music"> <input type="submit" value="submit"> </form> cant get it to send data from database to this page shoppingbasket.php |
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