populate drop down list from database using mysqli prepared statements
Question
As the title says i want to generated data from database and show it in dropdown. I made a code and it doesn't show any error but the thing is it echos the code in dropdown. It works when i just echo it. And one more thing is it doesn't show DISTINCT results.
Here is my code:
<html>
<head>
<title>FILTER</title>
</head>
<body>
<?php include 'conn.php';?>
<?php
$stmt = $con->prepare("SELECT DISTINCT author, book_name, language FROM bk_tst_fltr ");
$stmt->execute();
$stmt->bind_result($author,$book_name,$language);
$stmt->store_result();
echo "<select name='book'>";
while($row=$stmt->fetch()){?>
<p><?php echo '<option value="$row["author"]">"$row["author"]"</option>'; ?></p>
<?php }
echo "</select>";
?>
</body>
</html>
Is shows $row["author"] in the dropdown.
Can anyone solve this problem???
Thank You.
My DATABASE
id author book_name language price
1 Kishore One english 500
2 Kumar two english 600
3 Kishore three german 700
2 answers
solution1
1 ACCPTED 2013-07-11 07:34:02
You should use double quotes for variable expanding, or exit the string and concat normally.
Also, change to while($stmt->fetch()){?>
echo '<option value="'.$author.'">"'.$author.'"</option>';
Try also using this query instead:
"SELECT author, book_name, language FROM bk_tst_fltr GROUP BY author"
solution2
1 2013-07-11 07:43:41
The very statement of question is wrong.
Prepared statements has absolutely nothing to do with dropdown lists.
And never have to be.
What you have to populate is an array. Which later have to be used in a template.
So, divide your question into two.
- How to select some data into array.
- How to output an array in a template.
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