为什么 mysqli_num_rows() 需要期望参数 1 为 mysqli_result，在...中给出的布尔值？

Question

我在 wampserver 中的一个 php 文件中遇到问题。我有 3 个 php 文件，分别命名为 initial, registration and confirmation.

初始和注册 php 文件正在相应地工作。但是我的第三个文件确认是 not.My table 名称是 "user_information".

代码如下：

initial.php

<?php

$db_name = "myappdb";
$mysql_user = "root";
$mysql_pass = "";
$server_name = "localhost";

$con = mysqli_connect($server_name, $mysql_user, $mysql_pass, $db_name);
if (!$con) {
    echo"Error in connection try again..." . mysqli_connect_error();
} else {
    echo"<h3>Database is successfully connected!...</h3>";
}
?>

registration.php

<?php

require'initial.php';
$name = "angelina";
$user_name = "tom";
$user_pass = "cruise123";

$sql_query = "insert into user_information    values('$name','$user_name','$user_pass');";

if (mysqli_query($con, $sql_query)) {
    echo"<h3> Data is successfully inserted...</h3>";
} else {
    echo"Data Insertion is wrong..." . mysqli_error($con);
}
?>

confirmation.php

<?php

require'initial.php';
$user_name = "tom";
$user_pass = "cruise123";


$sql_query = "select name from user_informtion where user_name = '" . mysqli_real_escape_string($con, $user_name) . "' and user_pass =      '" . mysqli_real_escape_string($con, $user_pass) . "' ";
$result = mysqli_query($con, $sql_query);

if (mysqli_num_rows($result) > 0) {
    $row = mysqli_fetch_assoc($result);
    $name = $row["name"];
    echo"<h3>Hello welcome" . $name . "</h3>";
} else {
    echo"No information is available...";
}
?>

而且我不知道为什么会出现此错误：

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in.

请帮我解决这个问题。

Answer 1

您在 confirmation.php 页面上的查询有误。

您的查询：

 $sql_query="select name from user_informtion where user_name like'$user_name' and user_pass like'$user_pass';";

如果您想使用 like 子句，请将您的查询更改为：

 $sql_query="select name from user_informtion where user_name like '%$user_name%' and user_pass like '%$user_pass%'";

但我建议您使用 '=':

 $sql_query="select name from user_informtion where user_name ='".$user_name."' and user_pass = '".$user_pass."'";

Answer 2

使用 "=" 而不是 LIKE 并使用 mysqli_real_escape_string() 来防止 sql 注入。

替换：

$sql_query="select name from user_informtion where user_name like'$user_name' and user_pass like'$user_pass';";

有了这个：

$sql_query="select name from user_informtion where user_name = '".mysqli_real_escape_string($con, $user_name)."' and user_pass = '".mysqli_real_escape_string($con, $user_pass)."' ";

为什么 mysqli_num_rows() 需要期望参数 1 为 mysqli_result，在...中给出的布尔值？

Why does mysqli_num_rows() needs to expect parameter 1 to be mysqli_result,boolean given in...?

php

database

wampserver