Question

现在我的它显示即使我的数据库中已存在该名称，用户名也可用
PHP

session_start();
$conne=mysqli_connect("localhost","username","","my_db");

$name = $_POST['name'];

$check_if_exists="SELECT * FROM names WHERE name = '$name'";
$count = mysql_num_rows($check_if_exists);
if($count != 0) { 

 $_SESSION['msg']="The name already exists. Please try another name";

} else {

$sql="INSERT INTO names(name) values('$name')";
if ($conn->query($sql) === TRUE) {
   $_SESSION['msg']="The name was inserted successfully";
  }
}

header("location:form.php");

不知何故，它显示该名称可用，它将插入en给出消息，该行动是成功的
HTML

<div class="msg"> 
  <?php if(isset($_SESSION['msg']))
{
    echo $_SESSION['msg'];
    unset($_SESSION['msg]);
}
?>
</div>

<form action="check_insert.php" method="post">

<input type="text" name="name" id="name" class="form-control" required> 
<input class="btn btn-default" type="submit" value="Voeg domein toe!">

</form>

Answer 1

您的代码中存在一些问题：

$check_if_exists="SELECT * FROM names WHERE name = '$name'";
$count = mysqli_num_rows($check_if_exists);

这应该是那样的（你错过了mysqli_query函数）：

$check_if_exists="SELECT * FROM names WHERE name = '$name'";
$result = mysqli_query($check_if_exists);
$count = mysqli_num_rows($result);
if($count != 0) { 
//your code
}

这一行：

if ($conn->query($sql) === TRUE) 
{
    $_SESSION['msg']="The name was inserted successfully";
}

应该是这样的（你声明的连接变量为$ conne而不是$ conn ）：

if ($conne->query($sql) === TRUE) 
{
    $_SESSION['msg']="The name was inserted successfully";
}

旁注：

我不认为，如果你不在按钮中使用name="submit"它会起作用。

Answer 2

在您的代码中，您在致电mysqli_num_rows之前致电mysqli_query。 mysqli_num_rows计算mysqli_query返回的结果集上的行数，而不计算查询（文本）本身返回的行数。

显示该用户名即使已存在也可用

2 个答案: