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PODisplaying Data
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    copied!<p>I have a drop down list which is populated from a MYSQLi query. I want the user to select an option and the values/record associated pulled from the database and displayed to the user based upon FirstName.</p> <p>my table does not have an id column (I didnt make any. Starts with FirstName)</p> <p>It displays the table with column titles but it does not fetch the data dynamically from the db which happens to be coded in the while loop in [view code] below. <strong>I've spent my whole day and whole night trying to figure this out. Help :(</strong> </p> <p><strong>list code</strong></p> <pre><code>&lt;!DOCTYPE&gt; &lt;html&gt; &lt;head&gt; &lt;title&gt;Update Data&lt;/title&gt; &lt;/head&gt; &lt;body&gt; &lt;form name="form_update" method="post" action="update_test.php"&gt; &lt;?php $con=mysqli_connect("localhost","root","","ismat_db"); //============== check connection if(mysqli_errno($con)) { echo "Can't Connect to mySQL:".mysqli_connect_error(); } else { echo "Connected to mySQL&lt;/br&gt;"; } //This creates the drop down box echo "&lt;select name= 'FirstName'&gt;"; echo '&lt;option value=""&gt;'.'--- Please Select Person ---'.'&lt;/option&gt;'; //$query=mysqli_query($con,"SELECT id,FirstName FROM persons"); $query = mysqli_query($con,"SELECT FirstName FROM persons"); //$query_display = mysqli_query($con,"SELECT * FROM persons"); while($row=mysqli_fetch_array($query)) { echo "&lt;option value='". $row['id']."'&gt;".$row['FirstName'] .'&lt;/option&gt;'; } echo '&lt;/select&gt;'; ?&gt; &lt;input type="submit" name="submit" value="Submit"/&gt; &lt;/form&gt; &lt;br/&gt;&lt;br/&gt; &lt;a href="main.html"&gt; Go back to Main Page &lt;/a&gt; &lt;/body&gt; &lt;/html&gt; </code></pre> <p><strong>View Code</strong></p> <pre><code>&lt;!DOCTYPE&gt; &lt;html&gt; &lt;head&gt; &lt;title&gt;Update Data&lt;/title&gt; &lt;/head&gt; &lt;body&gt; &lt;?php $con=mysqli_connect("localhost","root","","ismat_db"); if(mysqli_errno($con)) { echo "Can't Connect to mySQL:".mysqli_connect_error(); } if(isset($_POST['FirstName'])) { $name = $_POST['FirstName']; $fetch="SELECT Firstname FROM persons WHERE Firstname = '".$name."'"; $result = mysqli_query($con,$fetch); if(!$result) { echo "Error:".(mysqli_error($con)); } //display the table echo '&lt;table border="1"&gt;'.'&lt;tr&gt;'.'&lt;td align="center"&gt;'. 'From Database'. '&lt;/td&gt;'.'&lt;/tr&gt;'; echo '&lt;tr&gt;'.'&lt;td&gt;'.'&lt;table border="1"&gt;'.'&lt;tr&gt;'.'&lt;td&gt;'.'First Name'.'&lt;/td&gt;'.'&lt;td&gt;'.' LastName'.'&lt;/td&gt;'.'&lt;td&gt;'. 'Gender' .'&lt;/td&gt;'.'&lt;td&gt;'. 'Subject'. '&lt;/td&gt;'.'&lt;td&gt;'.' Hobbies' .'&lt;/td&gt;'.'&lt;/tr&gt;'; //Supposed to collect data from db-I tried using _array,_assoc instead of _row //and got mysqli_result() // requires one parameter given boolean while($data=mysqli_fetch_row($result)) { echo ("&lt;tr&gt;&lt;td&gt;$data[0]&lt;/td&gt;&lt;td&gt;$data[1]&lt;/td&gt;&lt;td&gt;$data[2]&lt;/td&gt;&lt;td&gt;$data[3]&lt;/td&gt; &lt;td&gt;$data[4] &lt;/td&gt;&lt;/tr&gt;"); } echo '&lt;/table&gt;'.'&lt;/td&gt;'.'&lt;/tr&gt;'.'&lt;/table&gt;'; } ?&gt; </code></pre>
 

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