[英]Attempting to update results generated from a while loop
My main issue that I am running into is basically this: 我遇到的主要问题基本上是这样的:
I have a while loop that generates results from a query. 我有一个while循环,可从查询生成结果。 With the results that have been generated, I want the ability to update the table the original query was from. 有了生成的结果,我希望能够更新原始查询所来自的表。
The query produces the expected results, but the table is not being updated when I click the REMOVE button. 该查询产生了预期的结果,但是当我单击“删除”按钮时,该表没有被更新。 I am also trying to find a solution for the results to be updated after the UPDATE query executes... 我还试图为UPDATE查询执行后要更新的结果找到解决方案...
<?php
$sql = "SELECT * FROM vehicles WHERE sold='n' ORDER BY year DESC";
$query = mysql_query($sql);
while ($row = mysql_fetch_array($query)) {
echo
"
<tr>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['year'],"</td>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['make'],"</td>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['model'],"</td>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'><input type='submit' name='remove' value='REMOVE' style='background-color:#C33;color:white;padding:10px;border-radius:5px;width:70px'/></td>
</tr>";
if(isset($_POST['remove'])){
$removeSql = "UPDATE `table`.`vehicles` SET `display`='0' WHERE `vin`='{$row['vin']}'";
mysql_query($removeSql) or die('check that code dummy');
}
}
mysql_close($connection);
?>
That's a submit button, will not work without form tag. 这是一个提交按钮,没有表单标签将无法使用。 You can't do it this way. 你不能这样子。
You can write the remove code
on a separate page and convert that submit button
to normal button
and pass vin id
on click
of that button
and call
that page using ajax
. 您可以在单独的页面上编写remove code
,并将该submit button
转换为normal button
并在click
该button
传递vin id
, click
使用ajax
call
该页面。
Or if you don't know ajax
and want to do it on that page itself then do it this way : 或者,如果您不了解ajax
并想在该页面本身上进行操作,请按以下方式进行操作:
<?php
$sql = "SELECT * FROM vehicles WHERE sold='n' ORDER BY year DESC";
$query = mysql_query($sql);
while ($row = mysql_fetch_array($query)) {
echo
"
<tr>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['year'],"</td>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['make'],"</td>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['model'],"</td>
<td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>
<form action="" method="POST">
<input type="hidden" name="vin_id" value="<?php echo $row['vin']; ?>">
<input type='submit' name='remove' value='REMOVE' style='background-color:#C33;color:white;padding:10px;border-radius:5px;width:70px'/>
</form></td>
</tr>";
}
if(isset($_POST['remove'])){
$removeSql = "UPDATE `table`.`vehicles` SET `display`='0' WHERE `vin`='".$_POST['vin_id']."'";
mysql_query($removeSql) or die('check that code dummy');
}
mysql_close($connection);
?>
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