if(isset($_POST['name']))
{
$productID = $_POST['pid'];
$cat_id = $_POST['cat_id'];
$name = $_POST['name'];
$picture = $_POST['picture'];
$description = $_POST['description'];
$me_level = $_POST['me_level'];
$pe_level = $_POST['pe_level'];
$runs = $_POST['runs'];
$copy = $_POST['copy'];
$stock_level = $_POST['stock_level'];
$price = $_POST['price'];
//INSERT
$query2 = "UPDATE products SET cat_id =' $cat_id', name =' $name', picture =' $picture', description =' $description', me_level =' $me_level', pe_level =' $pe_level', run =' $runs', copy =' $copy', stock_level =' $stock_level', price =' $price' WHERE serial=$productID";
$result = mysql_query($query2);
if($result){
echo '<div class="alert alert-success">
<button type="button" class="close" data-dismiss="alert">x</button>
<strong>Success!</strong> Product has been updated.
</div>';
}
else{
echo "<div class='alert alert-error'>
<button type='button' class='close' data-dismiss='alert'>x</button>
<strong>Error!</strong> failed again ".$productID."
</div>";
}
};
我已經插入了一個工作表 - 工作正常 - 但我無法正常工作。用PHP不能正常工作更新MySQL
怎麼了您的列數據中的所有前導空格?你從數據庫中得到什麼錯誤(請搜索如何獲取)? – Mat
如果您使用的是mysql_ *函數,而不是mysqli或pdo,請不要像這樣插入$ _Posts ...使用「$ yourpost = mysql_real_escape_string($ _ POST [yourPost]);」 –
不要說某些「不起作用」,你應該更精確並描述1)你想要做什麼2)什麼具體失敗,3)你試圖解決它。 – slhck