Question

我还有其他几个相同的功能正常运行，但我无法使用最后一点（照片被'上传'不存储在数据库中）。任何帮助都将非常感激。 我理解mysql语句是一个已弃用的版本，将会更改。

Ajax功能

function uploadProLicense5() {
    var ajax = new XMLHttpRequest();
    var dataFive = new FormData();
    dataFive.append("professionalLicensePhotoFifthProivder", document.querySelector("#professionalLicensePhotoFifthProivder").files[0]);
    ajax.open("POST", "home.php");
    ajax.onreadystatechange = function() {
        if (ajax.readyState === 4) {
            if (ajax.status === 200) {
                // OK
                $('#success_prolicense5').fadeIn().html("Photo Saved");
                setTimeout(function(){
                    $('#success_prolicense5').fadeOut("Slow");
                }, 2000);
                // here you can use the result (ajax.responseText)
            } else {
                // not OK
                alert('failed to upload photo, please try again');
            }
        }
    };
    ajax.send(dataFive);
}

HTML

<div class="well col-lg-4 col-lg-offset-4 col-md-4 col-md-offset-4 col-sm-10 col-sm-offset-1 col-xs-10 col-xs-offset-1" id="professionalphotoRow5">
    <div class="text-center">
        <label>Upload professional license photo:</label>
    </div>
    <div class="text-center">
        <label id="label5thproviderPro">for 5th provider</label>
    </div>
    <div class="col-sm-offset-4 col-xs-offset-4">
        <input name="professionalLicensePhotoFifthProivder" id="professionalLicensePhotoFifthProivder" class="filestyle" data-input="false"
               type="file" accept="image/*">
    </div>
    <?php
    $savedTxt = "Photo Saved";
    $notSavedTxt = "Photo Not Saved";
    if (!empty($userRow[pro_license_photo_provider_five])) {
        echo "<i class='glyphicon glyphicon-ok col-sm-offset-4 col-xs-offset-4' style='color: greenyellow;'>" . "</i> " . $savedTxt;
    }
    ?>
    <br>
    <div class="col-sm-offset-4 col-xs-offset-4">
        <input id="saveProfessionalPhoto5" type="button" class="btn btn-default"
               name="saveProfessionalPhoto5" onclick="uploadProLicense5()"
               value="upload photo" style="background-color: #1d2b5c; color: white"/>
    </div>
    <span id="success_prolicense5" class="text-success col-sm-offset-4 col-xs-offset-4"></span>
</div>

PHP

$db_pro_license_img_provider_five = addslashes(file_get_contents($_FILES['professionalLicensePhotoFifthProivder']['tmp_name']));
$pro_image_data_provider_five = base64_encode($userRow[pro_license_photo_provider_five]);
$pro_data_provider_five = substr($pro_image_data_provider_five, strpos($pro_image_data_provider_five, ","));
if(!empty($db_pro_license_img_provider_five)){
    $sql = "UPDATE `wp_form` SET
    `pro_license_photo_provider_five`='$db_pro_license_img_provider_five'
    WHERE id =" . $_SESSION['user'];
    mysql_query($sql) or die(mysql_error());

    $res = mysql_query("SELECT * FROM wp_form WHERE id=" . $_SESSION['user']);
    $userRow = mysql_fetch_array($res);
}

这是我的网络请求

然后我的空洞:(

Ajax没有发送FormData

0 个答案: