Question

我正在尝试从php函数读取响应，但响应是未定义的值。但是，如果我使用chromes inspect元素并进入网络选项卡，我可以读取项目的正确值。

这里是代码：

Javascript功能

function chatHeartbeat(){       
$.ajax({        
url: 'chat.php?action=chatheartbeat',
type: 'post',
dataType: 'json',    
success: function(response) {       
    $.each(response.items, function(item){
        //alert(item.m);                
        $(".chatboxcontent").append('<div class=chatboxmessage><span class="chatboxmessagefrom">'+item.f+': </span><span class="chatboxmessagecontent">'+item.m+'</span></div>');
    });         
    setTimeout('chatHeartbeat();',3000);
},
error: function(){
    $(".chatboxcontent").append('<div class=chatboxmessage><span class="chatboxmessagefrom">ERROR!</span><span class="chatboxmessagecontent">UNKNOWN</span></div>');
    alert("failed");
}

});
}

这里是php代码：

function chatHeartbeat() {
if ($idchat!=""){
    $sql = "select * from chat where idchat=".$idchat." and recd=0 order by id ASC";                
    $result=$mysqli->query($sql);
    if ($result->num_rows>0){
        while($row=$result->fetch_array()){
            $sqlup="UPDATE chat SET recd = '1' WHERE idchat = ".$row['id'];
            if ($mysqli->query($sqlup)=== FALSE){
                echo "Error updating record" . $mysqli->error;
            }
            $item.=<<<EOD
                       {            
            "f": "{$row['from']}",
            "m": "{$row['message']}"
       },
EOD;
}

    }
}else{
    $item.=$item.=<<<EOD
                       {            
            "f": "{$mysqli->connect_errno}",
            "m": "{$mysqli->connect_error}"
       },
EOD;
}
    }
}else{
    $item.=$item.=<<<EOD
                       {            
            "f": "{$mysqli->connect_errno}",
            "m": "{$mysqli->connect_error}"
       },
EOD;
}
?>
{
    "items": [
          <?php echo $item;?>
    [
}
<?php
}
?>

Answer 1

如果您使用格式错误的json格式回复，则可能会发生这种情况。尝试简单地将数据放入数组/对象中，然后使用json_encode 在它上面：

<?php
$data = [];
$data["f"] = "value1";
$data["m"] = "value2";
echo json_encode($data);
?>

Answer 2

虽然没有它你通常很好，你可以并且应该设置Content-Type标题：

<?php
$data = array();
$files = array();

$allFiles = scandir($_REQUEST['dir_name']);
$files = array_diff($allFiles, array('.', '..'));
foreach($files as $key=>$file){
    $data[$key]['name'] = $file;
    if(is_dir(_REQUEST['dir_name'].'/'.$file)){
    //Get small image of folder here and store that in $data[$key]['preview']
        $data[$key]['type'] = 'dir';
    }
    else{
        $extension = substr(strrchr($filename, "."));
        $data[$key]['type'] = $extension;
        switch ()$extension) { 
        case 'jpeg' : {
            //Get small thumbnail of jpeg image here and store that in $data[$key]['preview']; 
        }
        case 'png' : {
            //Get small thumbnail of png image here and store that in $data[$key]['preview']; 
        }
        case 'pdf' : {
            //Get small thumbnail of pdf here and store that in $data[$key]['preview']; 
        }
        case 'rar' : {
            //Get small thumbnail of default .rar image here and store that in $data[$key]['preview']; 
        }
        }
    }
}
echo json_encode($data);
?>

注意：这是从另一篇文章中获取的（Returning JSON from a PHP Script）

jquery：无法读取php函数的响应

2 个答案: