CODEWITHSUNDEEP
phphtmlmysql
djac
djac

Reputation: 197

Populate drop down list from mysql in in php html

I am using the below code to populate drop down list in php html,

<?php                       
$mid="mario";
$sql = "SELECT * FROM tbl_prdy WHERE col_master_id = '$mid'";

$result = mysqli_query($conn,$sql);
echo "<select name='list'>";
while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['col_of_fa'] . "'>" . $row['col_of_fa'] . "
    </option>";
}
echo "</select>";
?>

But, I am getting internal server error. I have debugged the code and found that the issue is with the following 2 lines in the above code. There is not much information in server logs. Can you tell me what might be the issue with the following 2 lines of code?

while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['col_of_fa'] . "'>" . $row['col_of_fa'] . 
"/option>";
}

Upvotes: 0

Views: 56

Answers (3)

Pawan Thakur
Pawan Thakur

Reputation: 591

//Try this :

while ($row = mysqli_fetch_array($result)) { ?>
    <option value="<?php echo $row['col_of_fa'] ?>" ><?php echo $row['col_of_fa'] ?>
    </option>
<?php }

Upvotes: 0

A.A Noman
A.A Noman

Reputation: 5270

You would use this 

while($row=mysqli_fetch_assoc($result )){
}

or

while($row=mysqli_fetch_array($result )){
}

Upvotes: 0

Arun Kumaresh
Arun Kumaresh

Reputation: 6311

mixing mysqli with mysql

change

$row = mysql_fetch_array($result)

to

$row = mysqli_fetch_array($result)

Upvotes: 2

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