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POReally Easy PHP query. Displaying from separate PHP file
 

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  1. POReally Easy PHP query. Displaying from separate PHP file
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    <p>I'm a PHP n00b. I have this code, however it is on my main page. I assume like jQuery there must be a way to display this without having the actual script on the main document that the users access?</p> <p>Or is there a way I can do this tidier without echo'ing everything?</p> <pre><code> include('talentsearch.php'); while($row = mysqli_fetch_array($result2)) { echo "&lt;div class='talent_result_wrapper' data-experience='" . $row['divTagExp'] . "' data-salary='" . $row['divTagExp'] . "'&gt;"; echo "&lt;ul&gt;"; echo "&lt;li&gt;&lt;strong&gt;Talent ID: &lt;/strong&gt;" . $row['CandidateID'] . "&lt;/li&gt;"; echo "&lt;li&gt;&lt;strong&gt;Resides: &lt;/strong&gt;" . $row['Town'] . "&lt;/li&gt;"; echo "&lt;li&gt;&lt;strong&gt;Salary Required: &lt;/strong&gt;£" . $row['SalaryMin'] . "&lt;/li&gt;"; echo "&lt;li&gt;&lt;strong&gt;Experience: &lt;/strong&gt;" . $row['CandidateExperience'] . " Years &lt;/li&gt;"; echo "&lt;li&gt;&lt;strong&gt;Industy: &lt;/strong&gt;" . $row['PrimarySector'] . "&lt;/li&gt;"; echo "&lt;li&gt;&lt;strong&gt;Specialism: &lt;/strong&gt;" . $row['PrimarySector'] . "&lt;/li&gt;"; echo "&lt;/div&gt;"; } </code></pre> <p>I have tried putting it in a function like this:</p> <pre><code>function getTalent() { } </code></pre> <p>And then calling the function like this</p> <pre><code>&lt;?php include('talentsearch.php'); getTalent() ?&gt; </code></pre> <p>But I get an error whereas I don't if I just run the code normally.</p> <pre><code>Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in </code></pre> <p>So I assume I'm not doing it right.</p>
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    1. USMcam435
    1. USMcam435
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    1. COwhats in getTalent function?
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      1. POReally Easy PHP query. Displaying from separate PHP file
      1. UStonoslfx
    2. COWhere's `$result2` supposed to be defined? Currently it's `null` and that's what's causing the error.
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      1. POReally Easy PHP query. Displaying from separate PHP file
      1. USRolando Isidoro
    3. COWhatever you are passing to `mysqli_fetch_array()` does not exist inside the function. It is not global or not defined when you call the function.
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      1. POReally Easy PHP query. Displaying from separate PHP file
      1. USSumurai8
 

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