dongyun3335 2013-01-28 15:12
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使用mysqli_query进行ajax表单验证

so far I have used pure javascript to validate my form, but I need to add a mysqli query into the mix. Only I'm not so great with jquery and ajax. I can do a simple login form but this is a bit more complicated. can anyone give me any pointers on how I could add a jquery/ajax component to validate this:

foreach($_POST as $key=> $for) {

     if(!empty($for) && $key != 'send' && $key != 'title')  {

        $usercheck =  "SELECT email FROM users WHERE email = '$for'";
        $usercheck = $db->query($usercheck);

     if($usercheck->num_rows > 0) {$x="1"; continue;}
     if($usercheck->num_rows == 0){$x="2"; break;}
     }
  }

     if($x == "2") {$message = $for." is not a regestered email";}
     if($x == "1") {  // valid - submit.
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2条回答 默认 最新

  • dreljie602951 2013-01-28 15:47
    关注

    what you can do is to send $.post like this:

        $.post("test.php", { "post1": "something", "post2":"somethingelse" }, // those will be sent via post to test.php
      function(data){// the returned data
        console.log(data.return1); // here just logging to the console. **optional**
        console.log(data.return2); 
        // complete your process 
      }, "json"); // specifying the type as json also optional
    

    in your test.php

    foreach($_POST as $key=> $for) {
    
     if(!empty($for) && $key != 'send' && $key != 'title')  {
    
        $usercheck =  "SELECT email FROM users WHERE email = '$for'";
        $usercheck = $db->query($usercheck);
    
     if($usercheck->num_rows > 0) {$x="1"; continue;}
     if($usercheck->num_rows == 0){$x="2"; break;}
     }
      }
    
     if($x == "2") {$data['message'] = $for." is not a regestered email";
       echo json_encode($data); // echo to pass back to $.post .. json_encode() in case of using json
       }
     if($x == "1") {  // valid - submit
      $data['message'] = 'valid'; // pass the message as valid post
    echo json_encode($data); 
    }
    

    Remember:

    If you are posting a form submit to add event.preventDefault() to your javascript function to handle the form manually. here you can find more about it.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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