Question

在过去的几个小时里，我一直在尝试使用我的变量编写一个正确的插入语句。我应该使用表单来制作表格并将数据插入其中。我也是php的新手。我每次尝试都会遇到“没有插入”的消息。

我的PHP代码

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

    $table = $_POST['table'];
    $primarykey = $_POST['primarykey'];
    $field1 = $_POST['field1'];
    $field2 = $_POST['field2'];
    $field3 = $_POST['field3'];
    $fieldint = $_POST['fieldint'];
    $fieldint2 = $_POST['fieldint2'];
    $fieldint3 = $_POST['fieldint3'];

    $inprimarykey = $_POST['inprimarykey'];
    $infield1 = $_POST['infield1'];
    $infield2 = $_POST['infield2'];
    $infield3 = $_POST['infield3'];
    $infieldint = $_POST['infieldint'];
    $infieldint2 = $_POST['infieldint2'];
    $infieldint3 = $_POST['infieldint3'];

    mysql_query("INSERT INTO $table ($primarykey,$field1,$field2,$field3,$fieldint,$fieldint2,$fieldint3) VALUES ('$inprimarykey','$infield1','$infield2','$infield3','$infieldint','$infieldint2','$infieldint3')",$conn) or die ("NOTHING INSERTED");
}

Answer 1

首先，您不应再使用mysql_函数，而应使用mysqli_函数。

此外，对于您应使用mysqli_real_escape_string的任何恶意代码，您无法撤消用户输入。

对不起，我改写了你的代码。我使用类Mysqli来创建连接。如果出现错误，它将输出MySQL返回的输出：

    if (isset($_POST["INSERT"])) {
        $conn = new Mysqli( 'localhost', 'username', 'password', 'db-name' );
        $table=$conn->real_escape_string( $_POST['table'] );

        $data = array(
            $_POST['primarykey'] => $conn->real_escape_string( $_POST['inprimarykey'] ),
            $_POST['field1'] => $conn->real_escape_string( $_POST['infield1'] ),
            $_POST['field2'] => $conn->real_escape_string( $_POST['infield2'] ),
            $_POST['field3'] => $conn->real_escape_string( $_POST['infield3'] ),
            $_POST['fieldint'] => $conn->real_escape_string( $_POST['infieldint'] ),
            $_POST['fieldint2'] => $conn->real_escape_string( $_POST['infieldint2'] ),
            $_POST['fieldint3'] => $conn->real_escape_string( $_POST['infieldint3'] )
        ); 
        $query = sprintf( "INSERT INTO %s (`%s`) VALUES('%s');", $table, implode( '`,`', array_keys( $data ) ), implode( "','", $data ) );
        if( $conn->query( $query ) === false ) {
            printf("Error: %s\n", $conn->error);
        }
    }

如果这有助于你，请告诉我。

PHP MySql仅使用变量插入语句

我的PHP代码

1 个答案: