我正在尝试让网页从表中提取现有的sql记录，并使用表单来更新这些记录中的空列。我似乎无法更新这些记录以在表中实际更新。我一直在到处打猎，我尝试的一切都不起作用。我相信我遇到的问题是让它从表中获取 id，并使用它来更新日志表。我只是在学习PHP和SQL，所以完全有可能我忽略了一些非常愚蠢的东西。任何帮助都非常感谢。
$conn = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('there was a problem connecting to the database' . mysql_error());
$sql = "
SELECT Part
, Lot
, Qty
, AnodTemp
, Amp
, SealTemp
, PerformedBy
, DateTimePerformed
, FinalAnodThickness
, QtyPass
, CheckedBy
, DateTimeChecked
, id
FROM logs
";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$unapproved = $row['CheckedBy'];
if($unapproved == null) {
echo "<br><br><br> Part: " . $row['Part']. " / Lot: " . $row['Lot']. " / Qty: " . $row['Qty']. " / AnodTemp: " . $row['AnodTemp']. " / Amp: " . $row['Amp']. " / SealTemp: " . $row['SealTemp']. " / PerformedBy: " . $row['PerformedBy']. " / ID: " . $row['id']; ?>
<form action="index.php" method="post">
Final Anod Thickness:<br>
<input type="text" name="FinalAnodThickness">
<br><br>
Qty Pass:<br>
<input type="text" name="QtyPass">
<br><br>
<input type="submit" id="submit" value="Submit" name="submit">
<br><br>
</form>
_____________________________________________________________________
<?php
if (isset($_POST['submit'])){
$FinalAnodThickness= $_POST['FinalAnodThickness'];
$QtyPass= $_POST['QtyPass'];
$CheckedBy= $_SESSION['CheckedBy'];
$sql = "UPDATE logs SET FinalAnodThickness = $FinalAnodThickness WHERE EXISTS (SELECT id FROM logs)";
}
}
}
}
else {
echo "0 results";
}
$conn->close(); ?>
你只需要执行你拥有的更新语句：
$sql = "UPDATE logs SET FinalAnodThickness = $FinalAnodThickness WHERE EXISTS (SELECT id FROM logs)";
if ($conn->query($sql) === TRUE) {
echo "Records updated successfully";
} else {
echo "Error updating records: " . $conn->error;
}