错误 mysqli_fetch_array() 期望参数 1 为 mysqli_result，给定字符串

警告：mysqli_fetch_array() 期望参数 1 为 mysqli_result，给定字符串

这是我的代码，谁能告诉我出了什么问题吗？

$result ="SELECT * FROM report" ;
if(mysqli_query($cons, $result)) {
echo("
<div class='sc'>
<table  id='to1' width='90%' border='0' cellspaceing='1' cellpadding='8' align='center'>
 <tr bgcolor='#9B7272'>
<td > ID </td>
<td > first_name </td>
<td > last_name </td>
<td > phone </td>
<td > address </td>
<td > email </td>
<td > birthdate </td>
<td > gender </td>
<td > city </td>
<td > dr_name </td>

</tr>

");
while($row = mysqli_fetch_array($result))
{
$ID         =$row['ID']; 
$first_name =$row['first_name'];
$last_name  =$row['last_name'];
$phone      =$row['phone'];
$address    =$row['address'];
$email      =$row['email'];
$birthdate  =$row['birthdate'];
$gender     =$row['gender'];
$city       =$row['city'];
$dr_name    =$row['dr_name'];

echo "  <tr bgcolor='#C7B8B8'>

Problem

您缺少如何将参数传递给mysqli_fetch_array() http://php.net/manual/en/mysqli-result.fetch-array.php.

Solution

因此，这一行：

if(mysqli_query($cons, $result)) {

应该

if($res = mysqli_query($cons, $result)) { // assign the return value of mysqli_query to $res

^{(FWIW, I'd go with $res = mysqli_query($cons, $result); then do if($res) {.)}

然后做

while($row = mysqli_fetch_array($res)) // pass $res to mysqli_fetch_array instead of the query itself

Why?

你正在给予mysqli_fetch_array()- 作为一个论点 -string其中包含您的查询。事情不是这样的。你应该传递的返回值mysqli_query()反而。因此，你也可以这样写：while($row = mysqli_fetch_array(mysqli_query($cons, $result))) {}（但这不是建议，只是向您展示它是如何工作的）。