Question

我被困在这里。我一直在尝试上传一些照片，同时将相册的ID（值在表单中的隐藏形式）传递给处理上传的相同php脚本。但我不知道如何分别传递专辑ID这是代码。

的js

input.addEventListener("change", function (evt) {
        document.getElementById("response").innerHTML = "<img src='../assets/admin/images/loading.gif' />"
        var i = 0, len = this.files.length, img, reader, file;

    for ( ; i < len; i++ ) {
        file = this.files[i];

        if (!!file.type.match(/image.*/)) {
            if ( window.FileReader ) {
                reader = new FileReader();
                reader.onloadend = function (e) { 
                    showUploadedItem(e.target.result, file.fileName);
                };
                reader.readAsDataURL(file);
            }
            if (formdata) {
                formdata.append("images[]", file);
            }
        }   
    }

    if (formdata) {
        $.ajax({
            url: "../assets/admin/ajaxupload/upload.php",
            type: "POST",
            data: formdata,
            processData: false,
            contentType: false,
            success: function (res) {
                document.getElementById("response").innerHTML = res; 
            }
        });
    }

PHP

//how do i retrieve the given album id value that was passed.

foreach ($_FILES["images"]["error"] as $key => $error) {
    if ($error == UPLOAD_ERR_OK) {
        $name = $_FILES["images"]["name"][$key];
        move_uploaded_file( $_FILES["images"]["tmp_name"][$key], "../../uploads/pics/" .$_FILES['images']['name'][$key]);


    }
}
echo "<p>Successfully Uploaded Images</p>";

请尽快回复。

Answer 1

（对不起，我没有时间解释，但希望有所帮助）

在你的ajax php文件中

$errors = array();  // initialize empty error array
if (sizeof($errors) == 0) {
       ...
       if ($securimage->check($captcha) == false) {
            $errors['captcha_error'] = 'wrong code';
        }
 }
  if (sizeof($errors) == 0) {
        // no errors, send the form
        $return = array('error' => 0, 'message' => 'OK');
        die(json_encode($return));
    } else {
       $errmsg = '';
        foreach ($errors as $key => $error) {
            // set up error messages to display with each field
            $errmsg .= " - {$error}\n";
        }
        $return = array('error' => 1, 'message' => $errmsg);
        die(json_encode($return));
   }

你的js

url: 'ajax.php',
        type: "POST",
        data: $('#formID').serialize(),
        success: function(msg) {
            try {
                //                    alert( "Data Saved: " + msg );
                json = jQuery.parseJSON(msg);
                //                      alert(json.error)
                if (json.error == 0) {
                } catch(e) {
                alert("Sorry, there was an error parsing the json");
            }
        },
        error: function(msg) {
            alert("Ajax request failed");
        }

使用ajax将数据传递到php脚本

1 个答案: