php 警告：mysqli_fetch_array() 期望参数 1 是 mysqli_result，第 6 行给出的对象
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Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given on line 6
提问by Hello World
<?php
if (isset($_GET['hash'])&&!empty($_GET['hash'])){
$hash = $_GET['hash'];
$message_query = "SELECT from_id, message FROM message WHERE hash='$hash'";
$run_messages = mysqli_query($con,$message_query);
while($row_messages = mysqli_fetch_array($con,$run_messages)){
$form_id = $row_messages['from_id'];
$message = $row_messages['message'];
$user_query = "SELECT username FROM admins WHERE id='$from_id'";
$run_user = mysqli_fetch_array($con,$user_query);
$from_username = $run_user['username'];
echo "<p><strong>$from_username</strong></p></br>";
}
}else{
header('Location: messages.php');
}
?>
I'm getting this error message:
我收到此错误消息：
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given on line 6
警告：mysqli_fetch_array() 期望参数 1 是 mysqli_result，第 6 行给出的对象
Here's line 6 & as you can see I have already included the $conwhich is my database connection.
这是第 6 行 & 正如您所看到的，我已经包含了作为我的数据库连接的$con。
while($row_messages = mysqli_fetch_array($con,$run_messages)){
while($row_messages = mysqli_fetch_array($con,$run_messages)){
采纳答案by Saty
mysqli_fetch_array ( mysqli_result $result [, int $resulttype = MYSQLI_BOTH ] )
mysqli_fetch_array ( mysqli_result $result [, int $resulttype = MYSQLI_BOTH ] )
First parameter is Specifies a result set identifier returned by mysqli_query()and second parameter is result type
第一个参数是Specifies a result set identifier returned by mysqli_query()，第二个参数是result type
Remove connection as first parameter
删除连接作为第一个参数
It would be
这将是
$row_messages = mysqli_query($run_messages,MYSQLI_ASSOC);
Your code is open for sql injection better use bind statement
你的代码是开放的 sql 注入更好地使用绑定语句
http://php.net/manual/en/mysqli-stmt.bind-param.php
http://php.net/manual/en/mysqli-stmt.bind-param.php
To check error in your connection and query use
要检查连接和查询中的错误，请使用
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
if (!$mysqli->query("SET a=1")) {
printf("Errormessage: %s\n", $mysqli->error);
}
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/mysqli.error.php
In second query you use
在您使用的第二个查询中
$user_query = "SELECT username FROM admins WHERE id='$from_id'";
$user_query = "SELECT username FROM admins WHERE id='$from_id'";
TYPO here
打错字在这里
$form_id !=$from_id
Change this to
将此更改为
$user_query = "SELECT username FROM admins WHERE id=' $form_id'";
$user_query = "SELECT username FROM admins WHERE id=' $form_id'";
回答by Matt
<?php
if (isset($_GET['hash'])&&!empty($_GET['hash'])){
$hash = mysqli_escape_string($con, $_GET['hash']);
$message_query = "SELECT from_id, message FROM message WHERE hash='$hash'";
$run_messages = mysqli_query($con,$message_query);
while($row_messages = mysqli_fetch_array($run_messages, MYSQLI_ASSOC)){
$from_id = $row_messages['from_id'];
$message = $row_messages['message'];
$user_query = "SELECT username FROM admins WHERE id='$from_id'";
$query_run = mysqli_query($con, $user_query);
$run_user = mysqli_fetch_array($query_run, MYSQLI_ASSOC);
$from_username = $run_user['username'];
echo "<p><strong>$from_username</strong></p></br>";
}
}else{
header('Location: messages.php');
}
?>
Essentially mysqli_fetch_array has one required parameter which is the result of a queryand an optional query of the result type. $run_messagesis the result of the first query and will be used for the execution of mysql_fetch_arrayand MYSQLI_ASSOCis the optional type that you will be using so you can access the values in $row_messages like you do.
本质上 mysqli_fetch_array 有一个必需参数，它是查询的结果和结果类型的可选查询。 $run_messages是第一个查询的结果，将用于执行mysql_fetch_array和MYSQLI_ASSOC是您将使用的可选类型，因此您可以像访问 $row_messages 一样访问值。
Read mysqli_fetch_arrayfor more information.
阅读mysqli_fetch_array了解更多信息。
Also, please remember to escape user valued with mysqli_escape_string before allowing the data to be queried into the database. This reduces the chance of SQLi.
另外，请记住在允许将数据查询到数据库之前使用 mysqli_escape_string 转义用户值。这减少了 SQLi 的机会。
