我已经调用了来自SQL的数据,并通过编号 选择,然后我想使用“mysqli_fetch_array”行[1]中的数据做一些计算。缺少来自SQL的可变数据
但以某种方式在第14行$ row [1]无法获得计算的日期。
的index.php
<?php
$con = mysqli_connect('localhost','root','','test2');
$query = mysqli_query($con,"SELECT * FROM count") or
die(mysqli_error($con));
while($row = mysqli_fetch_array($query))
echo "$row[id]. $row[quantity] <a href='edit.php?edit=$row[id]'>Draw
Patrs<br />";
?>
edit.php
<?php
if(isset($_GET['edit']))
{
$id = $_GET['edit'];
$con = mysqli_connect('localhost','root','','test2');
$query = mysqli_query($con,"SELECT * FROM count WHERE id='$id'");
$row = mysqli_fetch_array($query);
}
if(isset($_POST['drawquantity']))
{
$drawquantity = $_POST['drawquantity'];
$newquantity = $row[1]-$drawquantity;
$id = $_POST['id'];
$sql = "UPDATE count SET quantity='$newquantity' WHERE id='$id'";
$con = mysqli_connect('localhost','root','','test2');
$res = mysqli_query($con,$sql)
or die("Could not update".mysql_error());
echo "<meta http-equiv='refresh' content='0;url=index.php'>";
}
?>
<form action="edit.php" method="POST">
How mant you will take away: <input type="text" name="drawquantity" value=""
placeholder="<?php echo $row[1]; ?>"><br />
<input type="hidden" name="id" value="<?php echo $row[0]; ?>">
<input type="submit" value=" Update "/>
</form>
最有可能当你提交'edit.php',该ID是不在那里,因此所选的行也不存在,我的猜测是你应该在你的表单中保留相同的url – Ghost
你的代码容易受到SQL注射。请学习使用[预先准备的语句](https://www.youtube.com/watch?v=nLinqtCfhKY)。 –