Question

我正在使用jquery在codeigniter中创建文件上传系统。但我在这方面遇到了一些问题。我正在获取文件名。但我不知道为什么它不在服务器上传文件。它没有在控制台给我任何回应。请告诉我，如果我遗失了什么。

点按钮上的我的Jquery代码

#include <stdio.h>
#include <stdlib.h>

void *func(int a)
{
    void *retVal = NULL;

    if (a==3)
    {
        retVal = malloc(sizeof(int));
        if (retVal != NULL)
        {
            *((int *)(retVal)) = 5;
        }
    }
    else if (a==4)
    {
        retVal = malloc(sizeof(int));
        if (retVal != NULL)
        {
            *((char *)(retVal)) = 'b';
        }
    }
    else
    {
        fprintf(stderr, "return value is NULL");
    }

    return retVal;
}

int main ()
{
    int *ptr_int = func(3);
    char *ptr_char = func(4);

    fprintf(stdout, "int value = %d\n", *ptr_int);
    fprintf(stdout, "char value = %c\n", *ptr_char);

    free(ptr_int);
    free(ptr_char);

    return 0;
}

这是我的控制器代码

 var post_data = $('#post').val();
        if (!$('#post').val()) {
            // console.log('hit');
            // alert($('input[type=file]').val());

        if (typeof FormData !== 'undefined') {

        // send the formData
        var formData = new FormData( $("#formname")[0] );

        $.ajax({
            url : 'Profile/uploadimg',  // Controller URL
            type : 'POST',
            data : formData,
            async : false,
            cache : false,
            contentType : false,
            processData : false,
            success : function(data) {
                console.log(data);
            }
        });

    } else {
       message("Your Browser Don't support FormData API! Use IE 10 or Above!");
    }

我的html表单

public function uploadimg()
    {
    $var = $_FILES ['fileUp'];
    //print_r($var);
     if($this->input->post('fileupload')) {

    $config['upload_path'] = 'upload'; 
    $config['file_name'] = $var;
    $config['overwrite'] = 'TRUE';
    $config["allowed_types"] = 'jpg|jpeg|png|gif';
    $config["max_size"] = '1024';
    $config["max_width"] = '400';
    $config["max_height"] = '400';
    $this->load->library('upload', $config);

    if(!$this->upload->do_upload()) {               
        $this->data['error'] = $this->upload->display_errors(); 

        print_r( 'error');
    } else { 
        print_r("success");                                      
    }

}

Answer 1

请尝试让我知道。你的ajax请求

$(document).on("submit",'#formname',function(event){
    var form = $('#formname').get(0); 
    var formData = new FormData(form);  
    $.ajax({
      type: "POST",
      url: "<?php echo site_url("Profile/uploadimg"); ?>",
      data: formData,
      dataType: 'jSon',
      xhr: function() {
            var myXhr = $.ajaxSettings.xhr();
            return myXhr;
      },
      cache:false,                    
      contentType: false,
      processData: false,
      success: function(result){
        console.log(result);
      }                       
    });
    event.preventDefault();
});

和控制器代码： -

public function uploadimg()
{
    $var = $_FILES ['fileUp'];
    $config['upload_path'] = 'upload'; 
    $config['overwrite'] = 'TRUE';
    $config["allowed_types"] = 'jpg|jpeg|png|gif';
    $config["max_size"] = '1024';
    $config["max_width"] = '400';
    $config["max_height"] = '400';
    $this->load->library('upload', $config);

    if(!$this->upload->do_upload('fileUp')) 
    {               
        $this->data['error'] = $this->upload->display_errors(); 
        echo json_encode(array("result"=>$this->data['error']));
        exit;
    } 
    else 
    { 
        echo json_encode(array("result"=>"Success"));
        exit;                                   
    }
}

Answer 2

将mimeType:"multipart/form-data"添加到您的ajax调用中，如：

    $.ajax({
        url : 'Profile/uploadimg',  // Controller URL
        type : 'POST',
        data : formData,
        mimeType:"multipart/form-data",
        async : false,
        cache : false,
        contentType : false,
        processData : false,
        success : function(data) {
            console.log(data);
        }
    });

还要确保您的表单标记具有属性enctype="multipart/form-data"。

评论该行：$config['file_name'] = $var;。您可以按如下方式获取上传的文件数据（在print_r("success");上面添加）：

$uploadFileData = $this->upload->data();
$uploadFileName = (isset($uploadFileData['file_name']) && trim($uploadFileData['file_name']))? $uploadFileData['file_name'] : '';

图像没有在php codeigniter上传，也没有得到回复

2 个答案: