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POSQL table comparison, build query and print filtered data into dropdown list
 

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    <p>I have a stored value from one table called user_id. I have another table as such:</p> <p>client_projects:</p> <pre><code>+------------------+-------------+-------------+-------------+ | project_codename | client_id_1 | client_id_2 | client_id_3 | +------------------+-------------+-------------+-------------+ | Alpha | 1 | 2 | 3 | +------------------+-------------+-------------+-------------+ | Beta | 2 | 1 | 0 | +------------------+-------------+-------------+-------------+ | Gamma | 3 | 1 | 0 | +------------------+-------------+-------------+-------------+ </code></pre> <p>If the user_id = (client_id_1 or client_id_2 or client_id_3) then I want to print the corresponding project_codenames in a dropdown list. This is the php I have but is not returning the right results. I have marketed the areas I am unsure of.</p> <p>PHP(updated and solved):</p> <pre><code>&lt;?php $con = mysqli_connect("localhost","****","****","db") or die("Connection error: " . mysqli_error($con)); //stored sample value from another table $user_id = "1"; //Build query: Not sure if this is correct format $query = "SELECT * FROM client_projects WHERE client_id_1 = $user_id OR client_id_2 = $user_id OR client_id_3 = $user_id"; $result = mysqli_query($con, $query) or die("Query error: " . mysqli_error($con)); //Drop down list echo '&lt;select id="Projects" class="input"&gt;'; echo '&lt;option value="" selected="selected" disabled="disabled"&gt;Choose a Project...&lt;/option&gt;'; // Loop through the query results, outputing specific options one by one // Not sure of the loop for the options while ($row = mysqli_fetch_array($query)) { echo '&lt;option value="'.$row['project_codename'].'"&gt;'.$row['project_codename'].'&lt;/option&gt;'; } echo '&lt;/select&gt;'; mysqli_close($con); ?&gt; </code></pre>
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    1. USRob
    1. USRob
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    1. COIn addition to the missing mysqli steps per Roman's answer, your query is using IN incorrectly. You should instead write it as client_id_1 = ? OR client_id_2 = ? OR client_id_3 = ?
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      1. USMichael J. Anderson
    2. COMichael is right. If You're not sure if Your query is correct, check mysqli_query()'s return value and optionally use [mysqli_error()](http://us3.php.net/mysqli_error) function to display complete error message the server sent You.
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      1. USRoman Hocke
    3. CO@MichaelJ.Anderson I've updated the code above and tested but still getting a syntax error in the $query. Am I not coding it right?
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      1. USRob
 

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