PHP function after the last mysql value of table

Question

I want to call a function printPager($max_pages,$pg); to show up after the last value in the table. I've failed with inserting exit() and end().

<?php
function catPosts($id) {
    $id = (int) $id;
    $pgsize=2;
    $pg=(is_numeric($_GET["p"]) ? $_GET["p"] : 1);
    $start=($pg-1)*$pgsize;
    $img_total=mysql_query("SELECT COUNT(1) FROM posts WHERE SubcategoryID = '$id'");
    $img_total=mysql_fetch_row($img_total);
    $img_total=$img_total[0];
    $max_pages=$img_total / $pgsize; 
    $max_pages=ceil($img_total/$pgsize);
    $query = mysql_query("SELECT * FROM posts WHERE SubcategoryID = '$id' LIMIT $start, $pgsize");
    while($post = mysql_fetch_array($query)) {      
        echo "<tr><td><h2><a href=\"viewArticle.php?id=" . $post['ID'] . "\">" . $post['Title'] . "</a> by " . $post['Author'] .  "</h2></td></tr><tr><td colspan=\"2\">" . $post['Summary'] . "</td><td>"  . $post['Date'] . "</td></tr>";
        echo exit($query);
        printPager($max_pages,$pg);
    }
}
?>

moe · Accepted Answer · 2012-01-17 00:49:48Z

1

Since php is a synchronous language, not async like javascript, you can easily return or call the next function after your while-block.

while($post = mysql_fetch_array($query)) {
     echo "<tr><td><h2><a href=\"viewArticle.php?id=" . $post['ID'] . "\">" . $post['Title'] . "</a> by " . $post['Author'] .  "</h2></td></tr><tr><td colspan=\"2\">" . $post['Summary'] . "</td><td>"  . $post['Date'] . "</td></tr>";
     //not working here    
     //echo exit($query);
}
printPager($max_pages,$pg);

The function exit will exit your entire php-runtime.

answered Jan 17, 2012 at 0:49

moe

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1 Comment

Victor Nik Over a year ago

Thanks, moe. I've tried it and it wouldn't work. The problem's appeared to be in <table> tags in my html - once I removed them, everything worked. Anyhow I've solved it now.

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