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POHow does a leftjoin work while using a initialized prepared statement for MySQLI?
 

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  1. POHow does a leftjoin work while using a initialized prepared statement for MySQLI?
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    <p>The user comes to this page from a different page where they click on a link and it is appended with a <code>?photo_id=</code> and then the id number. I want certain information to be available to the viewer when they arrive at this page.</p> <p>I want them to be able to see the photo, the photo name, and the photographers name. The first two are not a problem, because the <code>photo_id</code>, <code>photo_filename</code>, and <code>photo_title</code> are all in the same table. As soon as I try to get information about this photo that is on a different tabled called <code>photographers</code>, that's when I experience problems.</p> <p>Here's my code:</p> <pre><code>$sql = 'SELECT photos.photo_id, photos.photo_title, photos.photo_filename, photos.photographer_id, photographers.photographer_id, photographers.photographer_name FROM photos LEFT JOIN photographers ON photos.photographer_id = photographers.photographer_id WHERE photo_id = ?'; //initialize prepared statement $stmt = $conn-&gt;stmt_init(); if ($stmt-&gt;prepare($sql)) { $stmt-&gt;bind_param('i', $_GET['photo_id']); $stmt-&gt;bind_result($photo_id, $photo_title, $photo_filename, $photographer_id); $OK = $stmt-&gt;execute(); $stmt-&gt;fetch(); } </code></pre> <p>The first three variables(<code>$photo_id</code>, <code>$photo_title</code>, and <code>$photo_filename</code>) work fine and I can echo them out on my page, but the other variables that I added from the LEFTJOIN such as <code>photographers.photographer_id</code> and <code>photographers.photographer_name</code> which are from a different table, will not work. As it is know, the page totally breaks when I do add the extra column names on the SELECT line.</p> <p>I have a feeling it has something to with the amount of variables in the <code>bind_result()</code> function. I think they need to match the number of columns mentioned on the SELECT line, but I don't really want to create a variable for each one. </p> <p>Can anybody help me out?</p>
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