Question

我正在尝试从HTML select标签向MySQL插入一个值：

<select name="venuesend">
    <option value="37">LATINO SPORTS CLUB</option>
    <option value="38">RED HOOK VISION CENTER</option>
    <option value="39">ANTHOLOGY FILM ARCHIVES</option>
    <option value="40">HUMMINGBIRD STUDIOS</option>
    <option value="41">THE PRODUCERS CLUB</option>
</select>

因此，如果选择LATINO SPORTS CLUB，则应将37的值发送给MySQL。但是，目前发布到MySQL的值始终为3。所以，我猜这是因为某种原因被截断了。

这是我的PHP和HTML表单：

<?php

            //if "submit" is clicked
            if(isset($_POST['upload2'])) {


                //check if image files were uploaded
                foreach($_FILES['images']['error'] as $err){
                    switch ($err){
                        case UPLOAD_ERR_NO_FILE:
                            echo 'No file sent.';
                            exit;
                    }
                }

                //iterate through each image file uploaded
                for($x=0; $x<count($_FILES['images']['tmp_name']); $x++){


                    //THIS IS WHERE VENUESEND IS POSTED

                    $screeningID2 = $_POST['venuesend'][$x];
                    $file_name = $_FILES['images']['name'][$x];
                    $file_tmp = $_FILES['images']['tmp_name'][$x];

                    ....

                        //SQL statement and Posting
                        $sql2 = "INSERT INTO screeningImages (screeningId, imageURL) VALUES ('$screeningID2', '$file_name')";
                        mysqli_query($db, $sql2);
                    } else{
                        echo "Only images can be stored.";
                    }
                }
            }

        ?>  

        //HTML Form
        <form method="post" action="screenings-admin.php" enctype="multipart/form-data">


        //SELECT

        <select name="venuesend">
            <option value="37">LATINO SPORTS CLUB</option>
            <option value="38">RED HOOK VISION CENTER</option>
            <option value="39">ANTHOLOGY FILM ARCHIVES</option>
            <option value="40">HUMMINGBIRD STUDIOS</option>
            <option value="41">THE PRODUCERS CLUB</option>
        </select>


            //Images Upload
            <input type="file" id="imageUpload" name="images[]" multiple="" onchange="javascript:updateList()"/>

            //submit button
            <input type="submit" name="upload2" value="Upload" id="upload2">
        </form>

Answer 1

问题是您正在尝试访问字符串数组（$screeningID2 = $_POST['venuesend'][$x];），因为您正在调用<select>。

$_POST[id] POST通过$x传递选定的值;实际上你根本不需要引用$screeningID2 = $_POST['venuesend'];，只需要：

3

您收到$_POST['venuesend']因为37已经是字符串{ user_id : "user id", list_devices_token : [....]; }，您实际上是在接受字符串的第一个字符。< / p>

希望这有帮助！：）

Answer 2

代码的某些部分不清楚，但这应该从它的外观来解决问题。

$screeningID2 = $_POST['venuesend'][$x]

到

$screeningID2 = $_POST['venuesend];

最后使用[$x]，您试图存储/读取存储在变量中的整个字符串中的字符。尝试使用此代码并更新我们，以便我们查看是否可以修复它。

MySQL：从选择标记中插入值变为截断

2 个答案: