2016-02-22 117 views
0

我想在codeIgniter中使用ajax提交表单,但出现错误。在codeigniter中使用Ajax提交表单

这是我的视图代码;

<div class="form-group input-control"> 
    <p id="error1" style="display: none; color: green"><b>Registered Successfully.</b></p> 
    <p id="succ" style="display: none; color: red"><b>E-mail Alreay Registered.</b></p> 
    <button class="button" onclick="newsletter()"><span class="icon-envelop"></span></button> 
    <input type="text" class="form-control" placeholder="Your E-mail..." value="" id="newsletter" onblur="if (this.value == '') {this.value = 'Your E-mail...';}" onfocus="if(this.value == 'Your E-mail...') {this.value = '';}"> 
    <br> 
    <p id="error" style="display: none; color: red"><b>Please Enter E-mail.</b> 
     <p> 
      <p id="error2" style="display: none; color: red"><b>Error while request..</b> 
       <p> 
</div> 

这里是我的脚本:

<script> 
    function newsletter() { 
     var email = $("#newsletter").val(); 
     if (email == "") { 
      $("#error").css("display", "block"); 
     } else { 
      $.ajax({ 
       type: "post", 
       url: "<?php echo site_url();?>/shopit/index/newsletter", 
       cache: false, 
       data: 'search=' + email, 
       success: function(response) { 
        if (response == "succ") { 
         $("#succ").css("display", "block"); 
         $("#error").css("display", "none"); 
         $("#error1").css("display", "none"); 
         $("#error2").css("display", "none"); 
        } else { 
         $("#succ").css("display", "none"); 
         $("#error").css("display", "none"); 
         $("#error1").css("display", "block"); 
         $("#error2").css("display", "none"); 
        } 
       }, 
       error: function() { 
        $("#succ").css("display", "none"); 
        $("#error").css("display", "none"); 
        $("#error1").css("display", "none"); 
        $("#error2").css("display", "block"); 
       } 
      }); 
     } 

    } 
</script> 

,这是我的控制器功能的代码;

public function newsletter(){ 
    $search= $this->input->post('search'); 
    $result = $this->Index_model->newsletter($search); 
    echo $result; 
} 

,最后我的模型功能:

public function newsletter($search){ 
    $query = $this->db->get_where("shopit_newsletter", array('email'=>$search)); 
    $result = $query->result(); 

    if($query->num_rows() > 0){ 
      echo "error"; 
    }else{ 
     $data = array(
      "email"=>$search 
     ); 
     $this->db->insert("shopit_newsletter", $data); 
     echo "succ"; 
    } 
} 

这是抛出错误,并打算到错误,什么是错的这个代码,请帮帮我,我用Google搜索,但没有找到确切的结果。

这是控制台快照。

console snapshot

+3

你做了什么样的错误?显示控制台输出。 – StreetCoder

+0

传递给getElementById()的空字符串 –

+0

未清除您的评论。只需更多参考资料就可以正确解释显示控制台或错误输出的屏幕截图 – StreetCoder

回答

0

所有型号的功能首先应该是

public function newsletter($search) 
{ 
    $query = $this->db->get_where("shopit_newsletter", array('email'=>$search)); 
    $result = $query->result();  
    if($query->num_rows() > 0) 
    { 
     return "error"; 
    } 
    $data = array(
     "email"=>$search 
    ); 
    $this->db->insert("shopit_newsletter", $data); 
    return "succ";   
} 

控制器应

public function newsletter(){ 
    $search= $this->input->post('search'); 
    $result = $this->Index_model->newsletter($search); 
    echo json_encode(array("data"=>$result)); 
} 

而且在Ajax请求

success: function(response) { 
if (response.data == "succ") {