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[英]getting the value of a array from controller to view with ajax in codeigniter
[英]print value from view to controller codeigniter with ajax onchange without load page
我想使用條件視圖與視圖的CodeIgniter中的值而不加載頁面。 我試過跟蹤錯誤,我放棄了,我不知道,這是什么錯誤
jquery ajax:
var site = '<?php echo site_url();?>';
$(document).ready(function(){
$("#comboKec").change(function (){
var selectedText = $(this).find("option:selected").text();
var selectedValue = $(this).val();
//alert(selectedValue);
//$("#res").text(site);
$.ajax({
type: 'POST',
serviceUrl: site+'anggota/index',
data: {'kec' : selectedValue},
success: function(data) {
console.log(selectedValue);
},
error: function(msg) {
console.log('error : '+msg);
}
});
});
});
HTML:
<select>
<option value="aaaa">1</option>
<option value="bbbb">2</option>
</select>
控制器:
$kec = $this->input->post('kec');
if($kec="aaaa"){ echo "Data 1"; }
else if($kec="bbbb"){ echo "Data 2"; }
else { echo "No Data Selected"; }
我認為你需要改變兩個地方才能看到colsole的不同之處
$.ajax({
type: 'POST',
serviceUrl: site+'anggota/index',
data: {'kec' : selectedValue},
success: function(data) {
> console.log(selectedValue);
> console.log(selectedValue);
},
error: function(msg) {
console.log('error : '+msg);
}
});
並將名稱命名為select as
> <select name="comboKec">
<option value="aaaa">1</option>
<option value="bbbb">2</option>
</select>
哦,你的另一個錯誤是在$ kec =“aaaa”你應該使用$ kec ==“aaaa”!因為有了$ kec =“aaaa”你給copare分配價值你應該使用$ kec ==“aaaa”!
$kec = $_POST['kec'];
> if($kec=="aaaa"){ echo "Data 1" ;}
else if($kec=="bbbb"){ echo "Data 2"; }
else { echo "No Data Selected" ;}
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