我卡住了,我的代码没有错误,但报告的语法错误 - 我尝试了很多东西,无法让记录更新。
已成功连接 - 错误 - 您的SQL语法出错;检查与您的MySQL服务器版本对应的手册,以便在''10',weight'4.00',price '69.0000'附近使用正确的语法,其中products_id ='29''在第3行
<?php error_reporting(E_ALL);
ini_set('display_errors', '1');
?>
<?php
include 'connect.php';
if(isset($_POST['btn_submit'])){
$sql ="update products set
quantity'".$_POST['txt_quantity']."',
weight'".$_POST['txt_weight']."',
price'".$_POST['txt_price']."',
where products_id = '".$_POST['txt_id']."'
";
if(mysqli_query($con, $sql)){
header('Location:index.php');
}else{
echo "Error " .mysqli_error($con);
}
}
if(isset($_GET['id'])){
$sql = "SELECT products_id, products_quantity, products_weight,
products_price
FROM products
where products_id=" .$_GET['id'];
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result)>0){
($row = mysqli_fetch_assoc($result));
$id = $row['products_id'];
$quantity = $row['products_quantity'];
$weight = $row['products_weight'];
$price = $row['products_price'];
}
}
?>
<form action="" method="post">
<table>
<tr>
<td>Quantity</td>
<td><input name = "txt_quantity" value ="<?=$quantity?>"></td>
</tr>
<tr>
<td>Weight</td>
<td><input name = "txt_weight" value ="<?=$weight?>"></td>
</tr>
<tr>
<td>Price</td>
<td><input name = "txt_price" value ="<?=$price?>"></td>
</tr>
<tr>
<td><input type = "hidden" value ="<?=$id?>" name = "txt_id" ></td>
<td><input type="submit" name="btn_submit"></td>
</tr>
</table>
</form>
我认为错误是在where子句的某个地方,查看并尝试更改以下许多示例我无法将其发布。数据检索正常但我无法编辑然后发布更新。