插入查询语法错误 [英] insert query syntax error

问题描述

我无法找到错误..我求求你为我做这件事。代码无法提交数据

 include（'  ../的config.php'）; 
 
  if （isset（$ _ POST ['  submit']））
 {
  $ fullName  = $ _ POST ['  fullName']; 
  $ email  = $ _POST ['  email ]; 
  $ mobile  = $ _POST ['  mobile ]; 
  $ dob  = $ _POST ['  dob ]; 
  $ ddlType  = $ _POST ['  ddlType ]; 
  $ pincode  = $ _POST ['  pincode ]; 
  $ address  = $ _POST [' 地址]; 
  $ ddlState  = $ _POST ['  ddlState ]; 
  $ query  = $ _POST ['  query ]; 
  //    
  //   $ from = new DateTime（$ dob）;  
  //   $ to = new DateTime（'today'）;  
  //   echoage =。$ from-> diff（$ to） - > y;  
  //    
  //    
 
  echo   $ insertQuery  =   INSERT INTO`customerdetail`（fullName，email，dob，mobile ，地址，州，密码，InsurenceType，查询）VALUES（'$ fullName'，'$ email'，$ dob，$ mobile，'$ address'，'$ ddlState'，$ pincode，'$ ddlType'，'$ query ）; 
 
  $ insertCmd  = $ CS-> query（$ insertQuery）; 
 print_r（$ insertCmd）; 
 
 if（！$ insertCmd）{
  echo  ' < span class =code-string>数据未提交'; 
} 
 其他 {
  echo  ' 数据被提交'; 
} 
 
 
} 

解决方案

_POST [' submit']））
{

fullName =

_POST [' 全名'];

I am not able to find the error .. I beg you to do this for me. code is not able to submit data

include ('../config.php');

if (isset($_POST['submit']))
{
    $fullName =$_POST['fullName'];
    $email = $_POST['email'];
    $mobile = $_POST['mobile'];
    $dob = $_POST['dob'];
    $ddlType = $_POST['ddlType'];    
    $pincode = $_POST['pincode'];
    $address = $_POST['address'];
    $ddlState = $_POST['ddlState'];    
    $query = $_POST['query'];
//    
//    $from = new DateTime($dob);
//    $to   = new DateTime('today');
//    echo "age = ".$from->diff($to)->y; 
//
//    
    
    echo $insertQuery = "INSERT INTO `customerdetail`( fullName, email, dob, mobile, address, state, pincode, InsurenceType, query) VALUES ('$fullName','$email',$dob,$mobile,'$address','$ddlState',$pincode,'$ddlType', '$query')"; 
    
    $insertCmd = $CS->query($insertQuery);
    print_r($insertCmd);
    
    if(!$insertCmd){
        echo 'data is not submited';
    }
 else {
    echo 'data is  submited';
    }
    
    
}

解决方案

_POST['submit'])) {

fullName =

_POST['fullName'];

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