Question

我想用AJAX请求更新每条记录的数据库。问题是，当我提交编辑时，它给我一个500错误。更新＆＃39; RecordData＆＃39;给了我所有的数据，但就是这样。

我的更新文件。

    $data = $_POST['recordData'];
    str_parse($data, $data);

    echo   "Record Id is: " . $data[0];
    echo $query = "UPDATE bier 
              SET naam = '$data['naam'], brouwer = '$data['brouwer']', type = '$data['type']', gisting = '$data['gisting']', perc = '$data['perc']', inkoop_prijs = '$data['inkoop_prijs']'
              WHERE id= '".$data['id']."'";
    if (mysqli_query($conn, $query)) {
        echo "Record updated successfully";
    } else {
        echo "Error updating record: " . mysqli_error($conn);
    }
    mysqli_close($conn);

echo $_POST['recordId'];

我的AJAX功能

function editRecord(data){
console.log(data);

$.ajax({
    type: "POST",
    data: {
        database: 'updateRecords',
        recordData: data.serialize()
    },
    url: "actions/update.php",
    dataType: "html",
    async: false,
    success: function(data) {
        ModalClose();
    }
});
}

我的表格：

        <form id="FormBier" action="#" method="post">
        <table>
            <tr>
                <td>Id</td>
                <td><input type="text" name="id"></td>
            </tr>
            <tr>
                <td>Naam</td>
                <td><input type="text" name="naam"></td>
            </tr>
            <tr>
                <td>Brouwer</td>
                <td><input type="text" name="brouwer"></td>
            </tr>
            <tr>
                <td>Type</td>
                <td><input type="text" name="type"></td>
            </tr>
            <tr>
                <td>Gisting</td>
                <td><input type="text" name="gisting"></td>
            </tr>
            <tr>
                <td>Percentage</td>
                <td><input type="text" name="perc"></td>
            </tr>
            <tr>
                <td>Inkoop Prijs</td>
                <td><input type="text" name="inkoop_prijs"></td>
            </tr>
            <tr>
                <input type="submit" onclick="editRecord($('#FormBier'))" value="bewerk">
            </tr>
        </table>
    </form>
</div>

Answer 1

使用isset检查POST和变量，并按照下面的代码更改查询。

   echo $query = 'UPDATE bier SET naam = '.$data["naam"].', brouwer = '.$data["brouwer"].', type = '.$data["type"].', gisting = '.$data["gisting"].', perc = '.$data["perc"].', inkoop_prijs = '.$data["inkoop_prijs"].' WHERE id= '.$data["id"];

PHP＆amp; AJAX请求500（内部服务器错误）

1 个答案: