Trying to populate php dropdown with data from sql table
Question
I have been trying to populate a dropdown with a list of names from an mysql table. The table has columns for Name, Date, Host, and Info. I have looked around online, but I can't seem to find a solution to my issue. I am pretty sure it connects to the database, but I keep getting the error Notice: Undefined offset: 1 in C:\\xampp\\htdocs\\phpmsql\\delegates.php on line 16.
<?php
$db = new mysqli('127.0.0.1', 'root', '', 'munapp');
if($db->connect_errno) {
die('Sorry we having some connection problems');
}
$query = "SELECT Name FROM `conferences`";
$result2 = mysqli_query($db, $query);
$options = "";
while($row2 = mysqli_fetch_array($result2))
{
$options = $options."<option>$row2[1]</option>";
}
?>
<!DOCTYPE html>
<html>
<head>
<title> </title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<select>
<?php echo $options;?>
</select>
</body>
</html>
3 answers
solution1
0 2017-03-18 14:16:09
You wore trying to access the second column from the result, but you only selected one column.
Instead of the query:
$query = "SELECT Name FROM `conferences`";
select all columns like this:
$query = "SELECT name, date, host, info FROM `conferences`";
solution2
0 ACCPTED 2017-03-18 14:25:21
更改$options分配以从数组中的0索引填充：
$options = $options."<option>$row2[0]</option>";
solution3
0 2017-03-18 14:36:39
Like someone else pointed out, you need to access the first array element with an index of 0 instead of 1. Alternatively you could fetch your row like this:
while(row2 = mysqli_fetch_array($result2, MYSQLI_ASSOC)){
//accesss value here
}
which will give you the result as an associative array and then inside the loop access the values by their column namens like this:
$row2['Name']
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