我试图在php中创建一个数组,将通过ajax调用在javascript中解析为json。我注意到用$empty = array()
实例化的数组的第一个索引返回为{"0":[{..values here..}], "success":true}
。理想情况下,我将使用reply.data
和reply.success
访问它。的回复。成功的工作,但我似乎找不到如何创建一个数组没有0作为第一个索引。
我的php侧代码:
$result = array();
$record = array();
$resp = mysqli_query($this->conn, $sql);
if(mysqli_num_rows($resp)>0){
while($row = mysqli_fetch_assoc($resp)){
$record = array();
foreach($row as $key => $val){
$record[$key] = $val;
}
array_push($result,$record);
unset($record);
}
//return json_encode($result, JSON_PRETTY_PRINT);
return $result;
当我在javascript
中访问它 success: function(data){
data = JSON.parse(data);
if(data.success==true){ //able to access success with "data.success"
//there is an empty
$scope.conn = "connecting to room";
console.log(data.data.room_id); //does not work because index is 0
console.log(JSON.stringify(data));
返回什么
{"0":[{"room_id":"20","host_id":"","resp_id":"","offer":"","answerswer":"","status":"0"}],"success":true}
use this,
$result = array();
$resp = mysqli_query($this->conn, $sql);
if(mysqli_num_rows($resp)>0){
while($row = mysqli_fetch_assoc($resp)){
foreach($row as $key => $val){
$result["data"][$key] = $val
}
}
}
//return json_encode($result, JSON_PRETTY_PRINT);
return $result;
我认为你把这个过程弄得太复杂了。
foreach循环似乎是不必要的,因为您可以直接将$row加载到数组中。
public function xxx()
{
$result = array();
$resp = mysqli_query($this->conn, $sql);
if(mysqli_num_rows($resp)>0) {
$result['success'] = 'true';
while($row = mysqli_fetch_assoc($resp)){
$result['data'][] = $row;
}
} else {
$result['success'] = 'false';
}
return $result;
}
// instantiate object
//$obj = new Whatever the class is called()
echo json_encode($obj->xxx());