选择查询在php中不返回结果
我正在执行以下简单sql查询:选择查询在php中不返回结果,php,mysql,select,phpmyadmin,Php,Mysql,Select,Phpmyadmin,我正在执行以下简单sql查询: 从人员限制1中选择人员图片、人员性别,phpmyadmin中给出结果 我的php代码如下所示: $query = "SELECT people_picture, people_gender FROM people LIMIT 1"; $result = mysqli_query($con,$query); $row = mysql_fetch_assoc($result); $row1 = mysql_fetch_array($re
从人员限制1中选择人员图片、人员性别,phpmyadmin中给出结果
我的php代码如下所示:
$query = "SELECT people_picture, people_gender FROM people LIMIT 1";
$result = mysqli_query($con,$query);
$row = mysql_fetch_assoc($result);
$row1 = mysql_fetch_array($result);
print_r($row1);
print_r($row);
它没有给出一个错误,但它什么也不做 您混合了mysqli和mysql接口,这就是您陷入困境的原因:
请试试这个:
$con = mysqli_connect($host, $user, $pass, $dbase);
if(!$con){
echo 'Connection failed : '.mysqli_connect_error();
exit(0);
}
$query = "SELECT people_picture, people_gender FROM people LIMIT 1";
$result = mysqli_query($con,$query);
if(!$result){
echo 'Query failed : '.mysqli_error();
exit(0);
}
$row = mysqli_fetch_assoc($result); // mysql_fetch_assoc was the problem
print_r($row);
改为使用mysqli_fetch_assoc您是否正确连接到数据库?显示$con中的内容-也可以尝试向查询添加错误处理,以查看是否存在任何错误确定不要忘记添加故障检查,否则您可能会陷入调试噩梦:)。。。你很好。