我是学习如何设置数据库的初学者。 PHP脚本并按照示例 要做到这一点,那么当我运行login.php脚本时,我无法从数据库中检索数据, 我真的觉得这对其他人来说是一个非常简单的问题,但我试图解决它但是没有成功,那么有人可以看一下我的代码然后纠正它吗?
这是我的php脚本:
init.php:
<?php
$db_name = "webapp";
$mysql_username = "root";
$mysql_password = "";
$server_name = "localhost";
$con=mysqli_connect($server_name, $mysql_username, $mysql_password, $db_name);
if (!$con) {
echo "Connection Error ......." . mysqli_connect_error();
} else {
echo "<h3>Database connection Success .....</h3>";
}
?>
login.php:
<?php
require "init.php";
$user_name = "YASER";
$user_phone = "123456";
$sql_query = "select name from user_info where user_name like'$user_name'and
user_phone like'$user_phone';";
$result = mysqli_query($con,$sql_query);
if (mysqli_num_rows($result)>0)
{
$row = mysqli_fetch_assoc($result);
$name = $row["name"];
echo "<h3> Hello And Wellcome" . $name . "</h3>";
} else {
echo " No Info Is Available .......";
}
?>
答案 0 :(得分:0)
第一次:首先检查查询是executing
还是failing
if(!$result){ echo mysqli_error($con); }
第二名:使用=
代替like
$sql_query = "select name from user_info
where user_name='$user_name' and
user_phone='$user_phone'";
第3名:您需要在查询中正确的间距
like'$user_name'and
^^^ ^^^
到
like '$user_name' and
答案 1 :(得分:0)
您的查询中有错误。
尝试此以查找错误
$result = mysqli_query($con,$sql_query) or die(mysqli_error($con));
您的查询应如下所示......
'SELECT name FROM user_info WHERE user_name LIKE "'.$user_name.'" AND user_phone LIKE "'.$user_phone.'"';