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PHP无法更新数据库,提交后没有任何反应

转载 作者:行者123 更新时间:2023-11-29 11:17:19 24 4
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这是我的表单页面

<?php     $result= mysql_query("SELECT * FROM Staff WHERE StaffID = 1");

while ($myrow=mysql_fetch_assoc($result))
{

$PName=$myrow['PName'];
$PGender=$myrow['PGender'];
$PCoNum=$myrow['PCoNum'];
$PEmail=$myrow['PEmail'];
$PID=$myrow['PID'];
$PNat=$myrow['PNat'];
$PAdd=$myrow['PAdd'];
$PSal=$myrow['PSal'];
}
?>
<form action="SavePProfile.php" method="POST">
<div class="form-group">
<label for="PName" class="col-sm-2 control-label">Name: </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PName; ?>" id="PName">
</div>
</div>
<br></br>
<div class="form-group">
<label for="PGender" class="col-sm-2 control-label">Gender: </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PGender; ?>" id="PGender">
</div>
</div>
<br></br>
<div class="form-group">
<label for="PCoNum" class="col-sm-2 control-label">Contact No: </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PCoNum; ?>" id="PCoNum">
</div>
</div>
<br></br>
<div class="form-group">
<label for="PEmail" class="col-sm-2 control-label">Email Address: </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PEmail; ?>" id="PEmail">
</div>
</div>
<br></br>
<div class="form-group">
<label for="PID" class="col-sm-2 control-label">IC/Passport No: </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PID; ?>" id="PID">
</div>
</div>
<br></br>
<div class="form-group">
<label for="PNat" class="col-sm-2 control-label">Nationality: </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PNat; ?>" id="PNat">
</div>
</div>
<br></br>
<div class="form-inline">
<label for="PAdd" class="col-sm-2 control-label">Address: </label>
<textarea class="form-control" rows="5" cols="68" id="PAdd"><?php echo $PAdd; ?></textarea>
</div>
<br></br>
<div class="form-group">
<label for="PSal" class="col-sm-2 control-label">Salary (Monthly): </label>
<div class="col-sm-5">
<input type="text" class="form-control" value= "<?php echo $PSal; ?>" id="PSal">
</div>
</div>
<br></br>
<button type="submit" class="btn btn">Save</button>
<button type="submit" class="btn btn" onclick="location.href='HomePage.php';">Back</button>
</form>

虽然这是我的更新过程

  <?php    if (isset($_POST['PName']))
{
$PName=$_POST['PName'];
$PGender=$_POST['PGender'];
$PCoNum=$_POST['PCoNum'];
$PEmail=$_POST['PEmail'];
$PID=$_POST['PID'];
$PNat=$_POST['PNat'];
$PAdd=$_POST['PAdd'];
$PSal=$_POST['PSal'];




$UpdateProfile="UPDATE Staff
SET PName= '$PName', PGender= '$PGender', PCoNum= '$PCoNum', PEmail='$PEmail', PID='$PID', PNat='$PNat', PAdd='$PAdd', PSal='$PSal'
WHERE StaffID=1";

if (empty ($_POST['PName'])||empty ($_POST['PGender'])||empty ($_POST['PCoNum'])||empty ($_POST['PEmail'])||empty ($_POST['PID'])||empty ($_POST['PNat'])||empty ($_POST['PAdd'])||empty ($_POST['PSal']))
{
echo '<div class="callout callout-danger">
<h4>Notice!</h4>
<p>Unable to save!</p>
</div>';
}
else
{
mysql_query($UpdateProfile);

echo '<div class="callout callout-success">
<h4>Notice!</h4>
<p>User saved.</p>
</div>';
}


}
?>
<br></br>

数据能够显示在表单页面中,但是当我尝试单击“保存”按钮时,没有任何反应,没有通知没有错误,有人可以告诉我哪里出了问题吗?

最佳答案

您的输入都没有 name 属性,您应该更改:

<input type="text" class="form-control" value= "<?php echo $PName; ?>" id="PName">

与:

<input type="text" class="form-control" value= "<?php echo $PName; ?>" name="PName">

等等。

编辑:当然,您永远不应该使用 mysql_* 函数,如解释的 here .

关于PHP无法更新数据库,提交后没有任何反应,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39527703/

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