php 警告：mysqli_fetch_array() 期望参数 1 是 mysqli_result，对象在
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Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in
提问by age saputra
I'm got a little bit problem when I try to run this code.
当我尝试运行此代码时遇到了一些问题。
My problem is on line 32:
我的问题在第 32 行：
回答by Lorence Hernandez
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in
mysqli_fetch_array's 1st parameter must be a result of a query. what you are doing is you are you are passing the connection(which doesnt makes sense) and the query command itself.
mysqli_fetch_array 的第一个参数必须是查询的结果。你正在做的是你正在传递连接（这没有意义）和查询命令本身。
read the doc here:http://php.net/manual/en/mysqli-result.fetch-array.php
在这里阅读文档：http : //php.net/manual/en/mysqli-result.fetch-array.php
to fix this, execute the query first, then store the result to a variable then later fetch that variable.
要解决此问题，请先执行查询，然后将结果存储到变量中，然后再获取该变量。
$sql = "select * from privinsi";
$result = mysqli_query($connection,$sql);
while($r = mysqli_fetch_array($result)
{
/// your code here
}
回答by Davinder Kumar
You write sql query but didn't execute.
您编写了 sql 查询但没有执行。
$sql = "select * from privinsi";
$results = myqli_query($connection,$sql);
while($r = mysqli_fetch_array($results){
//enter your code here
}
