mysql结果无法更新值。请检查下面的代码和附图。我需要在单次提交时更新所有表格的值。我怎样才能做到这一点?foreach循环更新多个mysql值
<?php
if (isset($_POST['submit_update'])) {
$product_id = $_POST['id'];
// echo "<pre>";
// var_dump($product_id);die;
$p_name = $_POST['p_name'];
$cat_name = $_POST['cat_name'];
$stock_ava = $_POST['stock_ava'];
$max_odr = $_POST['max_odr'];
$price = $_POST['price'];
foreach ($product_id as $id) {
foreach ($p_name as $p_name_value) {
// echo "<pre>";
// var_dump($id);
$q = mysqli_query($conn, "UPDATE products SET product_name='$p_name_value', WHERE product_id='$id'");
if ($q==1) {
echo "success";
}else{
echo 'fail';
}
}
}
}
?>
如果测试的状态在这里,在这
,
是额外的WHERE
条款之前所有'mysqli_query()'调用和ec ho'mysqli_error($ con);'你会看到你的编译错误。然后你可以修复你自己的TYPO的 – RiggsFolly
你的脚本存在[SQL注入攻击]的风险(http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) 有看看发生了什么[小巴比表](http://bobby-tables.com/)即使 [如果你逃避投入,它不安全!](http://stackoverflow.com/questions/5741187/ sql -injection-that-gets-around-mysql-real-escape-string) 使用[prepared parameterized statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) – RiggsFolly