Question

我试图从数据库中删除一行，而且我正在使用以下代码。当我提交页面重新加载，如果我检查数据库，行仍然存在。没有错误，没有任何错误。我该怎么做才能解决这个问题？

HTML

<div class="delete_row">
    Sterge
    <form method="post" action="">
    *<input type="text" name="id_col" Placeholder="Id-ul coloanei"><br>
    <input type="submit" name="submit1" value="Sterge">
    </form>
</div>

PHP

$id_stergere=isset($_POST["id_col"]);
$submitcheck2=isset($_POST["submit1"]);

if($submitcheck2 && $id_stergere !==0 ){
    $sql = "DELETE FROM evenimente WHERE ID_even=$id_stergere";
    $result = query_mysql($sql);
}

Answer 1

query_mysql？

将此用于您的php：

error_reporting(E_ALL ^ E_NOTICE);

if(isset($_POST['submit1'])){

    $id_stergere=$_POST["id_col"];
    $submitcheck2=$_POST["submit1"];

    if($submitcheck2 && $id_stergere){
        $sql = "DELETE FROM evenimente WHERE ID_even=$id_stergere";
        $result = mysql_query($sql);
    }
}

你应该使用mysqli而不是。

使用此代替$sql = and $result = :

$link = mysqli_connect("localhost","db","password","user") or die("Error " . mysqli_error($link)); 
$query= "DELETE FROM evenimente WHERE ID_even=$id_stergere";
$result= mysqli_query($link,$query);

要获取数字，请使用：

$numrows = $result->num_rows;

对于fetch_array：

while($row = $result->fetch_array()){

      $var= $row['field'];
}

更新：将e rror_reporting(E_ALL ^ E_NOTICE);添加到php脚本的顶部。

Answer 2

if (isset($_POST["id_col"]))
    $id_stergere=$_POST["id_col"];

if (isset($_POST["submit1"]))
    $submitcheck2=$_POST["submit1"];

//additional check:
// if (!is_numeric($id_stergere)) die('there was a problem');
// if (!$submitcheck2) die('there was a problem');

    $sql = "DELETE FROM evenimente WHERE ID_even=".$id_stergere;
    $result = mysql_query($sql);

php query_mysql不返回任何内容

2 个答案: