php 警告：mysqli_fetch_array() 期望参数 1 是 mysqli_result，给定的字符串
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Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given
提问by emayne_designer
The problem i'm having is from the 4th line of code listed below. I get an error that says
我遇到的问题来自下面列出的第 4 行代码。我收到一个错误提示
Warning:
mysqli_fetch_array()expects parameter 1 to bemysqli_result,stringgiven
警告：
mysqli_fetch_array()期望参数 1 为mysqli_result，string给定
I don't have the variable enclosed in " "or ' 'so I'm not sure where the string recognition is coming from at this point. Can tell me how to fix this error?
我没有变封闭在" "或' '所以我不知道在哪里的字符串识别从在这一点到来。能告诉我如何解决这个错误吗？
$query = "SELECT * FROM questions WHERE id='question' LIMIT 5";
$result = mysqli_query($connection, $query);
if($query === FALSE) { die(mysql_error()); }
while($row = mysqli_fetch_array($query)){
$id = $row['id'];
$thisQuestion = $row['question'];
$question_id = $row['question_id'];
$q = '<h2>'.$thisQuestion.'</h2>';
$query2 = "SELECT * FROM answers WHERE question_id='$question' ORDER BY rand() LIMIT 5";
while($row2 = mysqli_fetch_array($query2)){
//...
}
}
回答by Marcin Nabia?ek
You have:
你有：
mysqli_fetch_array($query)
Should be:
应该：
mysqli_fetch_array($result)
Also in line 3 you have:
同样在第 3 行中，您有：
if($query === FALSE) { die(mysql_error()); }
Should be rather:
应该是：
if ($result === FALSE) { die(mysql_error()); }
