How do I display the same MySQL query twice using PHP?
Question
I selected values from my table, and wanted to display them two separate times. In the first instance, I wanted to display all cells in a table, and this works. In the second instance, I wanted to make a drop down from one of the columns. Only the first one works; how can I get both to work? I also need all of the rows in the table to display, which is why I thought a while loop was a good idea. But please correct me if I'm wrong in my thinking. Thanks.
<!DOCTYPE html>
<?php
$host = "localhost";
$username = "redacted";
$password = "redacted";
$db = "redacted";
$connection = mysqli_connect($host, $username, $password, $db);
if (mysqli_connect_errno()) {
echo "Connection error";
}
$query = "SELECT * FROM main";
$result = $connection->query($query);
?>
<html>
<head>
<title>Test Page</title>
</head>
<body>
<h1>Test Main Table Display</h1>
<table>
<?php
// first instance
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['user'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['count'] . "</td>";
echo "<td>" . $row['action'] . "</td>";
echo "<td><button>Expand</button></td>";
echo "</tr>";
}
?>
</table>
<h1>Test User Entry Appendment</h1>
<form>
<label>Select a user, or add a new user.</label><br>
<select>
<?php
// second instance
while($row = mysqli_fetch_array($result)) {
echo "<option>" . $row['user'] . "</option>";
}
?>
</select>
</form>
</body>
</html>
1 answers
solution1
0 ACCPTED 2018-11-03 15:40:46
add this line before second loop set the pointer back to the beginning
mysqli_data_seek($result, 0);
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