Question

我从特定ID的文件夹中获取图像列表。现在我正在获取文件名，但我也希望获得上传路径。

如何在一个函数中获取两个数据。

Jquery代码：

listFilesOnServer(project_id);

    function listFilesOnServer (project_id) {
        var items = [];
        uploadURI = uploadURI+'/'+project_id;
        console.log(project_id+'--KAL--'+uploadURI);
        $.getJSON(uploadURI ,function(data,path) {
            console.log(data);
            $('div #list-group').html("").html(items.join(""));
        });
    }

控制器代码：

function listFiles() {
        $this->load->helper('file');
        $project_id = $this->uri->segment(3);
        $builders_id = $this->admin_model->getBuilderID($project_id);
        $UPLD_PATH = $this->admin_model->builder_UPLD_PATH($builders_id);
        $upload_path = "./application/assets/images/" . $UPLD_PATH;
        $files = get_filenames($upload_path);
        echo json_encode($files);
    }

Answer 1

您应该修改控制器操作，使其返回json_encode(array('files'=>$yourFiles, 'filePath'=>$yourFilePath) );，如下所示：

function listFiles() {
        $this->load->helper('file');
        $project_id = $this->uri->segment(3);
        $builders_id = $this->admin_model->getBuilderID($project_id);
        $UPLD_PATH = $this->admin_model->builder_UPLD_PATH($builders_id);
        $upload_path = "./application/assets/images/" . $UPLD_PATH;
        $files = get_filenames($upload_path);
        echo json_encode(array('files'=>$files, 'uploadPath'=>$upload_path) );
        exit();
}

然后修改你的jquery代码来处理json响应并提取响应，如下所示：

listFilesOnServer(project_id);

function listFilesOnServer (project_id) {
    var items = [];
    uploadURI = uploadURI+'/'+project_id;
    console.log(project_id+'--KAL--'+uploadURI);
    $.getJSON(uploadURI ,function(data,path) {
        //Your upload path
        console.info("UPLOAD PATH: "+data.uploadPath);
        //Your files
        console.log(data.files);
        //Your processing logic goes here
        $('div #list-group').html("").html(items.join(""));
    });
}

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1 个答案: