javascript - 使用ajax mysql php上传图片

Question

我想尝试使用 php 和 mysql 上传图像。我正在使用表单通过 ajax 发送数据。

我的 HTML 代码:

<input type="file" name="logo" id="logo" class="styled">
<textarea rows="5" cols="5" name="desc" id="desc" class="form-control"></textarea>
<input type="submit" value="Add" id="btnSubmit" class="btn btn-primary">

Ajax 代码:

var formData = new FormData($("#frm_data")[0]);
$("#btnSubmit").attr('value', 'Please Wait...');
$.ajax({
    url: 'submit_job.php',  
    data: formData,
    cache: false,
    contentType:false,
    processData:false,
    type: 'post',
    success: function(response)

我的 php 代码 (submit_job.php):

$desc =  mysqli_real_escape_string($con, $_POST['desc']);
$date = date('Y-m-d H:i:s');
$target_dir = "jobimg/";
$target_file = $target_dir . basename($_FILES["logo"]["name"]);
move_uploaded_file($_FILES["logo"]["tmp_name"], $target_file);

Answer 1

试试这个:

J查询:

$('#upload').on('click', function() {
        var file_data = $('#pic').prop('files')[0];
        var form_data = new FormData();  // Create a FormData object
        form_data.append('file', file_data);  // Append all element in FormData  object

        $.ajax({
                url         : 'upload.php',     // point to server-side PHP script 
                dataType    : 'text',           // what to expect back from the PHP script, if anything
                cache       : false,
                contentType : false,
                processData : false,
                data        : form_data,                         
                type        : 'post',
                success     : function(output){
                    alert(output);              // display response from the PHP script, if any
                }
         });
         $('#pic').val('');                     /* Clear the input type file */
    });

PHP:

<?php
    if ( $_FILES['file']['error'] > 0 ){
        echo 'Error: ' . $_FILES['file']['error'] . '<br>';
    }
    else {
        if(move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']))
        {
            echo "File Uploaded Successfully";
        }
    }

?>