Question

我有php文件，它返回json对象。如果错误我会像这样设置它们，

if(move_uploaded_file($_FILES['file']['tmp_name'][$position], $uploaddir.$name)){
    $uploaded[] = array(
        'name' => $name,
        'file' => 'assets/uploads/'.$name
    );
}
else
{
    $error[] =  array('error' => $_FILES['file']['tmp_name'][$position].' file uploading failed');  
}

如果设置了错误，我需要显示它们。否则我想显示数据。到目前为止，我试过这个。但它不起作用

var displayUploads = function(data){
    var uploads = document.getElementById('uploads'),
                    anchor,
                    x;

    console.log(data);

    if(data.file === undefined)
    {
        for(x=0;x<data.length;x=x+1)
        {
             console.log(data[x].error;
        }      
    }
    else
    {
        for(x=0;x<data.length;x=x+1)
        {
            anchor = document.createElement('a');  
            anchor.href = '<?php echo base_url(); ?>'+ data[x].file;
            anchor.innerText = data[x].name;
            anchor.target = '_blank';

            uploads.appendChild(anchor); 
        }
    }  
}

Answer 1

尝试替换此行：

if(data.file === undefined)

用这个：

if(typeof(data.file) === 'undefined')

Json字符串，如果键错误设置显示。否则显示数据

1 个答案: