提交表单后如何检查数据库中存在的文本框值

Question

I have this HTML Code in a form, I would like to check whether this value that user input is already exist in the database or not.

<tr height="50">
    <td valign="top"><label for="CINumber"><span style="color:red">*</span>CI Name:</label></td>
     <td><input type="text" id="CINumber" name="CINumber"></td>                     
</tr>

I have this ``JavaScript function for sumbmit the form values:

function submit(){
    var cNumber = document.getElementById("CINumber").value;
     //many form value within this function but I showed this value only cause this value need to check when click submit button.
    $.ajax({
        url: 'PHPMethodCalls_AL.php',
        type: 'post',
        data: {'action':'AddNewCI', 'cNumber' : cNumber },
        success: function(data,status) {
            //alert(data);                                                              
            if(status == "success"){                            
                alert("CI Name can be assigned");
                alert("Successfully Register New CI Record.");                              
            } else {
                alert("CI Name is already present!");
                return ;
            }   
        },                              
        error: function(xhr, desc, err) {
            console.log(xhr);
            console.log("Details: " + desc + "\nError:" + err);
        }
    });                                                                         
}

And I call the PHPMethod Call in the 'PHPMethodCalls_AL.php' url..

if($_POST['action'] == "AddNewCI"){
    $cNumber = $_POST['cNumber'];
    $newCI = AddNewCI($cNumber);
    echo json_encode($newCI );
}

After that I call the PHP function that connect with the database.

function AddNewCI($cNumber){
    global $DB;
    if(!empty($cNumber)){   
        $namequery="SELECT Name from Item where Name='$cNumber'";
        $namecheck=mysql_query($namequery);
        $row=mysql_num_rows($namecheck);

        if($row==0){                     
            echo "success";
            $strSql1 ="INSERT INTO Item (Name)VALUES('$cNumber')";
            $res1 = $DB->Query($strSql1, false, $err_mess.__LINE__);                    
        }
        else {  
            echo "!success";
        }
    }                                                       
}

After click submit, I only got the (status == "success") then pop up 2 alert box even the data is already exist in the database. However, the duplicate value don't save in the database cause I made the UNIQUE for this fields. Please suggest me how should I do to get the the right alert box if the data was exist, show "Alredy exist" if not " show the success alert box"...

Answer 1

Change following code ...

                            url: 'PHPMethodCalls_AL.php',
                            type: 'post',
                            data: {'action':'AddNewCI', 'cNumber' : cNumber },

                            success: function(data) {

                            if(data == "success"){                            
                                alert("CI Name can be assigned");
                                alert("Successfully Register New CI Record.");                              
                            }

no need to use json_encode , you can directly echo any value if you want.

if($_POST['action'] == "AddNewCI"){
$cNumber = $_POST['cNumber'];
$newCI = AddNewCI($cNumber);
echo $newCI;
}

change you function return value...

function AddNewCI($cNumber){
    global $DB;

        if(!empty($cNumber)){   

            $namequery="SELECT Name from Item where Name='$cNumber'";
            $namecheck=mysql_query($namequery);
            $row=mysql_num_rows($namecheck);
        if($row==0){                     
                $strSql1 ="INSERT INTO Item (Name)VALUES('$cNumber')";
                $res1 = $DB->Query($strSql1, false, $err_mess.__LINE__); 
                return "success";                   
            }
        else
            {   
                 return "error";
            }

        }                                                       
   }

Answer 2

Change your success function with this in ajax request :

success:fucntion(data){
  if(data=="success"){
    alert("Your success message here.");
  }else{
    alert("Alert message whatever you want.");
  }
},

Answer 3

try this one less code and working well
here no need to display error message if you use jquery validation you can simply use "render" and render function displayed error message like required in jquery // When the browser is ready... $(function() { // Setup form validation on the #register-form element $("#addfrm").validate({

   // Specify the validation rules
    rules: {
      CINumber: {
            required: true,
            remote: {
            url: "PHPMethodCalls_AL.php",
            type: "post",
            data: {
              CINumber: function() {
                return $( "#CINumber" ).val();
              }
            }
          }
        }
    },

    // Specify the validation error messages
    messages: {
        CINumber: {
            required: "Please enter your CINumber",
            remote:"Entered CINumber already exists please enter another one"
       }
    },

    submitHandler: function(form) {
        form.submit();
    }
});
});
</script>

 PHPMethodCalls_AL.php
   this file return true when cinumber not exits on database and when it`s return false that means cinumber already exits on database      


<?php
   //add database configuration 
   $CINumber=$_POST['CINumber'];
   $query="SELECT * FROM `table_name` where `CINumber`=$CINumber";
   $result=mysql_query($query);
   $number_of_rows = mysql_num_rows($result);
   if($number_of_rows){
      echo "false";
   }else{
     echo "true";
   }
   die();
   ?>