mysqli_num_rows（）期望参数1为mysqli_result，第12行的C：\ wamp \ www \ DbConnect.php中给出了字符串

Question

这是我的PHP代码，每当我运行它我得到错误...数据库是在localhost wamp服务器上

<?php
$username ="root";
$password= "";
$hostname ="localhost";
$database ="mydata";
$check="";
$conn = new mysqli($hostname, $username, $password, $database);
$i = "SELECT * FROM `patient` WHERE 1";
$num_rows = mysqli_num_rows($i);
while($row = mysqli_fetch_array($i))
{
$r[]=$row;
$check=$row["emailid"];
}
if($check==NULL)
{
$r[$num_rows]="Record is not available";
print(json_encode($r));
}
else
{
$r[$num_rows]="success";
print(json_encode($r));
}
//mysql_close($conn);
?>

请为此建议解决方案。

Answer 1

改变你的联系

$conn=mysqli_connect($hostname, $username, $password, $database);

你有错误的SQL

改变这一点

$i = "SELECT * FROM `patient` WHERE 1";

至

$i = "SELECT * FROM `patient` WHERE your_cloumn_name = 1";

而你还没有执行你的

加

$result=mysqli_query($conn,$i);

然后

while($row = mysqli_fetch_array($result))