$str = 'http://example.com/en/search';
$part = explode('/',$str,5);
if ($part[3] == 'en') {
$url = $part[0].'//'.$part[2].'/'.$part[4];
} else {
$url = $str;
}
echo '$url:['.$url.']'; // http://example.com/search
点击这是我的按钮 -
<button id="survey_act" method="post" class="tiny ui blue button" type="button" value="<?php echo $surv['id']; ?>" >Activate Survey</button>
这是我的javascript -
<script>
$(document).ready(function(){
$(document).on("click","#survey_act", function(){
alert(this.value);
idx = this.value;
$.ajax({
type: "POST",
url: "<?php echo base_url('index.php/admin/survey/act_surveyby_id/')?>/"+idx,
}).done(function(msg){
if(msg=="success"){
alert('You Successfully Activated the Survey!');
}
});
});
});
</script>
这是我的控制器 -
public function act_surveyby_id($id){
$this->load->model('survey_m');
if($this->survey_m->insert_activate($id)){
echo "success";
}else{
echo "invalid";
}
}
这是我的模特 -
问题:当我点击激活调查时,它不会更改/更新调查的详细信息。我真的非常需要帮助。谢谢 。 。
答案 0 :(得分:0)
更改$ .ajax函数,如下所示
$.ajax({
url: '<?php echo base_url(); ?>index.php/admin/survey/act_surveyby_id',
type: "POST",
data: {
idx : idx;
},
和下面的控制器
public function act_surveyby_id(){
$id=$_POST['idx'];
$this->load->model('survey_m');
if($this->survey_m->insert_activate($id))
{
echo "success";
}else{
echo "invalid";
}
}