Home > Backend Development > PHP Tutorial > PHP中使用mysqli_fetch_object的一个错误

PHP中使用mysqli_fetch_object的一个错误

WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWB
Release: 2016-06-06 20:34:18
Original
1039 people have browsed it

<code> mysqli_select_db($con,"my_2db"); //选择操作库
    $query='SELECT password FROM user WHERE account='.$account;//定义sql查询语句
    $result=mysqli_query($con,$query); //发送sql查询

    if($obj=mysqli_fetch_object($result))//取查询完的结果
</code>
Copy after login
Copy after login

数据库中只有一条记录

PHP中使用mysqli_fetch_object的一个错误
当我查询account='13'时,按我php代码中返回的密码错误的提示,当我查询account='1a'时报错
报错如下

PHP中使用mysqli_fetch_object的一个错误
请问下是我数据库有问题还是其他一些原因,我在phpmyadmin中使用语句

<code>SELECT password FROM user WHERE account='13'
</code>
Copy after login
Copy after login

<code>SELECT password FROM user WHERE account='1a'
</code>
Copy after login
Copy after login

都是相同的结果,并未查询上的不妥
PHP中使用mysqli_fetch_object的一个错误

PHP中使用mysqli_fetch_object的一个错误

回复内容:

<code> mysqli_select_db($con,"my_2db"); //选择操作库
    $query='SELECT password FROM user WHERE account='.$account;//定义sql查询语句
    $result=mysqli_query($con,$query); //发送sql查询

    if($obj=mysqli_fetch_object($result))//取查询完的结果
</code>
Copy after login
Copy after login

数据库中只有一条记录

PHP中使用mysqli_fetch_object的一个错误
当我查询account='13'时,按我php代码中返回的密码错误的提示,当我查询account='1a'时报错
报错如下

PHP中使用mysqli_fetch_object的一个错误
请问下是我数据库有问题还是其他一些原因,我在phpmyadmin中使用语句

<code>SELECT password FROM user WHERE account='13'
</code>
Copy after login
Copy after login

<code>SELECT password FROM user WHERE account='1a'
</code>
Copy after login
Copy after login

都是相同的结果,并未查询上的不妥
PHP中使用mysqli_fetch_object的一个错误

PHP中使用mysqli_fetch_object的一个错误

发现哪里错了,查询语句错了,这句,

<code> $query='SELECT password FROM user WHERE account='.$account;
</code>
Copy after login

当含有字母时要注意是字符的写法

<code> $query='SELECT password FROM user WHERE account='.'"'.$account.'"';
</code>
Copy after login

有一种情况:mysqli_query查询account='1a'时,查询是失败的,此时mysqli_query返回一个布尔值false,而mysqli_fetch_object的第一个参数是一个mysqli_result 对象

你可以执行查询后,var_dump($result)看看查询结果

Related labels:
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Issues
php data acquisition?
From 1970-01-01 08:00:00
0
0
0
PHP extension intl
From 1970-01-01 08:00:00
0
0
0
How to learn php well
From 1970-01-01 08:00:00
0
0
0
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template