php,mysql - query data from mysql table and display results using for each

Question

Basically I first query specific columns from a mysql table and want to echo them all using foreach() in php. Here is my code.

$sql = "SELECT id, name, roll_no, email, title, abstract, supervisor, sub_date
        FROM student_data";

$result = $conn->query($sql);
$array = $result->fetch_assoc();

foreach ($array as $x => $value)
{
    echo $array["id"] . " " . $array["name"] . " " . $array["roll_no"] . " " . $array["title"] . " " . $array["supervisor"] . " " . $array["sub_date"] ."<br><br>";
}

It simply keeps repeating the first entry.

John Conde · Accepted Answer · 2015-04-03 16:10:19Z

2

You're only looping through the elements in the first result set. You need to iterate through the entire result set:

while($array= $result->fetch_assoc()) {
    echo $array["id"] . " " . $array["name"] . " " . $array["roll_no"] . " " . $array["title"] . " " . $array["supervisor"] . " " . $array["sub_date"] ."<br><br>";
}

answered Apr 3, 2015 at 16:10

John Conde

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Leonid Mednikov · Accepted Answer · 2015-05-15 11:43:25Z

0

Your main missunderstanding is that $array = $result->fetch_assoc(); returns only one result. To get the second one you should call $result->fetch_assoc() once again. So you should iterate in this way

while ($array = $result->fetch_assoc()) {
echo $array["id"] . " " . $array["name"] . " " . $array["roll_no"] . " " . $array["title"] . " " . $array["supervisor"] . " " . $array["sub_date"] ."<br><br>";
}

edited May 15, 2015 at 11:43

answered Apr 3, 2015 at 16:15

Leonid Mednikov

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