Question

我有一个数据库，其中有NOT NULL和NULL个字段（NULL属于VARCHAR类型）。
当我尝试通过查询在NULL字段中输入数据时，它不会插入它们数据不会同时输入：

使用表单我将数据插入NOT NULL字段
使用其他表单在NULL字段中插入数据。

为什么在NULL字段中输入数据的查询没有？我试图找到类似问题的答案，但它们不起作用或不适合我的问题：

MySQL Insert Select - NOT NULL fields
Insert NULL into DATE field MySQL 5.6

FIRST FORM register.php

<?php

   if($_SERVER['REQUEST_METHOD']=='POST'){
  // echo $_SERVER["DOCUMENT_ROOT"];  // /home1/demonuts/public_html
//including the database connection file
       include_once("config.php");

    $id_akR = $_POST['id_akR'];
    $numero_telefonoR = $_POST['numero_telefonoR'];


     if($id_akR == '' || $numero_telefonoR == '' ){
            echo json_encode(array( "status" => "false","message" => "Parameter missing!") );
     }else{

            $query= "SELECT * FROM RistoratoreAK WHERE id_akR='$id_akR' OR numero_telefonoR ='$numero_telefonoR' ";
            $result= mysqli_query($con, $query);

            $query2 = "SELECT ak_id, numero_telefono FROM AccountKit WHERE ak_id = '$id_akR' OR numero_telefono = '$numero_telefonoR'";
            $result2= mysqli_query($con, $query2);

            if(mysqli_num_rows($result) > 0){  
               echo json_encode(array( "status" => "false","message" => "User already exist in Ristoratore!") );
            }else if(mysqli_num_rows($result2) > 0) {
            echo json_encode(array( "status" => "false","message" => "User already exist in Cliente!") );

            }else{ 
             $query = "INSERT INTO RistoratoreAK (id_akR, numero_telefonoR) VALUES ('$id_akR','$numero_telefonoR')";
             if(mysqli_query($con,$query)){

                 $query= "SELECT * FROM RistoratoreAK WHERE numero_telefonoR ='$numero_telefonoR'";
                         $result= mysqli_query($con, $query);
                     $emparray = array();
                         if(mysqli_num_rows($result) > 0){  
                         while ($row = mysqli_fetch_assoc($result)) {
                                     $emparray[] = $row;
                                   }
                         }
                echo json_encode(array( "status" => "true","message" => "Successfully registered!" , "data" => $emparray) );
             }else{
                 echo json_encode(array( "status" => "false","message" => "Error occured, please try again!") );
            }
        }
                mysqli_close($con);
     }
     } else{
            echo json_encode(array( "status" => "false","message" => "Error occured, please try again!") );
    }

 ?>

第二种形式register2.php

<?php
if($_SERVER['REQUEST_METHOD']=='POST'){

include 'config2R.php'; 

 $con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);

 $nome = $_POST['nomeR'];
 $cognome = $_POST['cognomeR'];
 $data_nascita = $_POST['data_nascitaR'];
 $sesso = $_POST['sessoR'];
 $nome_ristorante = $_POST['nome_ristoranteR'];


 $CheckSQL = "SELECT nome_ristorante FROM RistoratoreAK WHERE nome_ristorante='$nome_ristorante'";

 $check = mysqli_fetch_array(mysqli_query($con,$CheckSQL));

 if(isset($check)){

 echo 'Ristorante già registrato';

 }
else{ 
$Sql_Query = "INSERT INTO RistoratoreAK (nomeR,cognomeR,data_nascitaR,sessoR,nome_ristorante) values ('$nome','$cognome','$data_nascita','$sesso','$nome_ristorante')";

 if(mysqli_query($con,$Sql_Query))
{
 echo 'Registration Successfully';
}
else
{
 echo 'Something went wrong';
 }
 }
}
 mysqli_close($con);
?>

我的数据库包含一个名为“RistoratoreAK”的表，字段为：

 id                   INT PrimaryKey
 id_ak                VARCHAR NOT NULL
 number               VARCHAR NOT NULL
 nomeR                VARCHAR NULL
 cognomeR             VARCHAR NULL
 sessoR               VARCHAR NULL
 data_nascitaR        VARCHAR NULL
 nome_ristorante      VARCHAR NULL

注意：对不起，如果代码不安全（我没有使用PDO），这段代码只是一个学习如何将数据上传到数据库的测试。

Answer 1

在第一个表单之后，使用id和id_ak将新条目插入到表中。这很好，而且很有效。

但在第二个表单之后，您不应该插入另一个条目，而是更新现有的条目（您之前创建的条目）。
要更新它，您需要知道现有条目的ID。

有了这个，您可以像这样进行UPDATE查询：

UPDATE 
    RistoratoreAK
SET
    nomeR = '$nome',
    cognomeR = '$cognome',
    data_nascitaR = '$data_nascita',
    sessoR = '$sesso',
    nome_ristorante = '$nome_ristorante'
WHERE 
    id = $existing_id

如何在NULL字段中插入VARCHAR数据

1 个答案: