我的代码可以从mysql表中检索数据。但是我无法更新检索到的数据。我相信代码是不言自明的。
<?php
session_start();
if($_SESSION['app'])
{
if(isset($_POST['submit']))
{
$acadexamo = $_POST['acadexamo'];
$acaddivio = $_POST['acaddivio'];
$acadpco = $_POST['acadpco'];
$acadyearo = $_POST['acadyearo'];
$acadspeciao = $_POST['acadspeciao'];
$acadboardo = $_POST['acadboardo'];
$db = mysqli_connect('localhost','root','','test');
$sqlz = mysqli_query($db, "
SELECT id
FROM recru_quali
WHERE app='$_SESSION[app]'");
$rowa = mysqli_fetch_assoc($sqlz);
$maxqua = count($acadexamo);
for($i=0;$i<$maxqua;$i++)
{
$sqla = mysqli_query($db, "
UPDATE recru_quali
SET acadexam='$acadexamo[$i]',
acaddivi='$acaddivio[$i]',
acadpc='$acadpco[$i]',
acadyear='$acadyearo[$i]',
acadspecia='$acadspeciao[$i]',
acadboard='$acadboardo[$i]'
WHERE id='$rowa[id]'
AND app = '$_SESSION[app]'");
}
}
?>
<form method="post" action="" enctype="multipart/form-data">
<div class="tab"><h3 align="center">Academic Qualifications:</h3>
<table id="employee_tablezz" align="center" border="1" width="100%">
<div id="form_div">
<?php
$db = mysqli_connect('localhost','root','','test');
$sql = mysqli_query($db, "SELECT * FROM recru_quali WHERE
app='$_SESSION[app]'");
$rows = mysqli_fetch_assoc($sql);
while($row = mysqli_fetch_assoc($sql))
{
?>
<tr><td><input type="text" name="acadexamo[]" value="<?php echo
$row[acadexam]; ?>"></td>
<td><input type="text" name="acaddivio[]" value="<?php echo
$row[acaddivi]; ?>"></td>
<td><input type="text" name="acadpco[]" value="<?php echo $row[acadpc]; ?
>"></td>
<td><input type="text" name="acadyearo[]" value="<?php echo
$row[acadyear]; ?>"></td>
<td><input type="text" name="acadspeciao[]" value="<?php echo
$row[acadspecia]; ?>"></td>
<td><input type="text" name="acadboardo[]" value="<?php echo
$row[acadboard]; ?>"></td></tr>
</table>
<button name="submit">Submit</button>
</form>
<?php
}
?>
上面的代码仅更新一行。请告诉我如何用相同的应用程序号更新所有行。我添加了图像。这些行已从数据库中检索出来并显示在输入中,以便用户可以对其进行修改。但不会。