Question

我有一个带有mysql的php，如果没有相同的信息，会将一些数据插入数据库

<?php 

$id=$_POST['id'];
$guildname=$_POST['guildname'];
$level=$_POST['level'];
$score=$_POST['score'];
$guildmaster=$_POST['guildmaster'];


$con = mysql_connect("localhost", "root", "");
if (!$con)
  {die('Could not connect to mysql: ' . mysql_error());} 

$mydb = mysql_select_db("gunbound");
if (!$mydb)
  {die('Could not connect to database: ' . mysql_error());} 

  $dup = mysql_query("SELECT Id FROM guildrequest WHERE Id='".$_POST['id']."'");
        if(mysql_num_rows($dup) >= 1){
            echo '<b>You have already ask for guild request.</b>';
        }
        else
        {

     $dup2 = mysql_query("INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')");
        }
 Print "<center>You have requested to join the guild.</center>"; 

mysql_close($con);
?>

但如果没有等于

的记录，则不会将记录添加到数据库中

也不执行此操作：

$dup2 = mysql_query("INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')");

即使代码：

if(mysql_num_rows($dup) >= 1){

说他可以做插入

的动作

请帮帮我

Answer 1

试试这个：

<?php 

$id=$_POST['id'];
$guildname=$_POST['guildname'];
$level=$_POST['level'];
$score=$_POST['score'];
$guildmaster=$_POST['guildmaster'];


$con = mysql_connect("localhost", "root", "");
if (!$con)
  {die('Could not connect to mysql: ' . mysql_error());} 

$mydb = mysql_select_db("gunbound");
if (!$mydb)
  {die('Could not connect to database: ' . mysql_error());} 

  $dup = mysql_query("SELECT Id FROM guildrequest WHERE Id='".$_POST['id']."'");
        if(mysql_num_rows($dup) >= 1){
            echo '<b>You have already ask for guild request.</b>';
        }
        else
        {

     $dup2 = mysql_query("INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')");
     return $dup2;
        }
 Print "<center>You have requested to join the guild.</center>"; 

mysql_close($con);
?>

我已将return添加到您的else语句中。我将执行你的$ dub2变量。如果你想保留变量$ dub2，那么你可以在中间执行你的查询。另一种方法是使用mysql_execute（）函数。

这将是MYSQLI的等价物：

<?php 

$id=$_POST['id'];
$guildname=$_POST['guildname'];
$level=$_POST['level'];
$score=$_POST['score'];
$guildmaster=$_POST['guildmaster'];

$host = "hostname";
$user = "username";
$password = "password";
$database = "database";

$con = mysqli_connect($host, $user, $password, $database);
if (!$con)
  {die('Could not connect to mysql: ' . mysql_error());} 


 $dup = "SELECT Id FROM guildrequest WHERE Id='".$_POST['id']."'";

mysqli_query($con, $dup);
  if (!$dup)
  {die('Could not connect to database: ' . mysql_error());} 

        if(mysqli_num_rows($dup) >= 1){
            echo '<b>You have already ask for guild request.</b>';
        }
        else
        {
     $dup2 = "INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')";
     mysqli_query($con, $dup2);
        }
 Print "<center>You have requested to join the guild.</center>"; 

mysqli_close($con);
?>

PHP MySQL插入

1 个答案: