4
我有一個內部多選的表單。在提交時我有這樣的功能:jquery serializearray以JSON多選形式
$("#addTrainForm").submit(function(event) {
var dataString = $(this).serializeArray();
console.log(dataString);
event.preventDefault();
$.ajax({
type: "POST",
url: "/index.php/trainings/insertTraining",
data: dataString,
dataType: 'json',
success: function(data) {
$("div#addModal").hide();
location.reload();
}
});
return false;
});
現在,在這裏你可以看到控制檯日誌,我得到如下:
0: Object
name: "date"
value: "14/10/2013 16:22:46"
__proto__: Object
1: Object
name: "pool"
value: "1"
__proto__: Object
2: Object
name: "repeat_0"
value: "1"
__proto__: Object
3: Object
name: "distance_0"
value: "1"
__proto__: Object
4: Object
name: "style_0"
value: "1"
__proto__: Object
5: Object
name: "change_0"
value: "1"
__proto__: Object
6: Object
name: "time_0"
value: "1"
__proto__: Object
7: Object
name: "options_0"
value: "1"
__proto__: Object
8: Object
name: "options_0"
value: "2"
__proto__: Object
9: Object
name: "repeat_1"
value: "2"
__proto__: Object
10: Object
name: "distance_1"
value: "2"
__proto__: Object
11: Object
name: "style_1"
value: "1"
__proto__: Object
12: Object
name: "change_1"
value: "2"
__proto__: Object
13: Object
name: "time_1"
value: "2"
__proto__: Object
14: Object
name: "options_1"
value: "4"
__proto__: Object
15: Object
name: "options_1"
value: "5"
__proto__: Object
16: Object
name: "options_1"
value: "6"
在那裏你可以看到options_0和options_1有多個值。
現在在得到被稱爲PHP函數我讀了所有的數據後,它straght寫入文件:
public function insertTraining(){
$data=$this->input->post();
$this->baselib->logIt(print_R($data,true));
}
在那裏我options_0
和options_1只有可用的最後一個值。我期待更多的值的數組或一個逗號的字符串或東西。這是我的日誌上的文本文件看起來像:
Array
(
[date] => 14/10/2013 16:22:46
[pool] => 1
[repeat_0] => 1
[distance_0] => 1
[style_0] => 1
[change_0] => 1
[time_0] => 1
[options_0] => 2
[repeat_1] => 2
[distance_1] => 2
[style_1] => 1
[change_1] => 2
[time_1] => 2
[options_1] => 6
)
我怎樣才能得到options_0和options_1所有值設置爲我的PHP腳本?有沒有更好的方式來使用我已經是我的功能(如可選參數?)或者我必須運行一個循環周圍的所有值,並找到選項值,並將它們放入一個數組,然後發送到PHP與JSON?
希望我設法解釋自己, 謝謝 詹姆斯
嘗試var_dump($ _ POST []); – Hackerman
同樣的結果。我跑了var_dump($ this-> input-> post());因爲codeIgniter似乎不喜歡我使用$ _POST [] – James
和var_dump($ _ POST); ? – Hackerman