php mysql_query() 期望参数 1 是字符串，给定资源
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mysql_query() expects parameter 1 to be string, resource given
提问by AphexBeirut
so my code is this..
所以我的代码是这样的..
<?php
$password=(!isset($_POST['password']));
$username=(!isset($_POST['username']));
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);
$query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
$result = mysql_query($query);
var_dump($result);
$num_rows = $result->$num_rows;
if ($num_rows)
{
echo "username already exist";
}
else
{
$query = "INSERT INTO tb_funcionario (nome_funcionario, username, password) VALUES (
'$_POST[nome_funcionario]',
'$_POST[username]',
'$_POST[password]'
)";;
$result = mysql_query($query) or die (mysql_error());
}
mysql_query($query);
mysql_close($bd_con);
?>
And its always giving me the "mysql_query() expects parameter 1 to be string, resource given" and i cant figure out how to solve it.
它总是给我“mysql_query() 期望参数 1 是字符串，资源给定”，我不知道如何解决它。
Can yall help me?
你能帮我吗？
回答by David
This is your problem:
这是你的问题：
$query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
$result = mysql_query($query);
You're running a query on the first line, which returns a "resource" as a result of the query. Then on the immediate next line you try to use that resource as another query to run again. You don't need the second line, the $resultcan be set in the first line.
您在第一行运行查询，该查询返回“资源”作为查询结果。然后在紧接的下一行，您尝试将该资源用作另一个查询以再次运行。你不需要第二行，$result可以在第一行设置。
回答by miku
You have some unnecessary/wrong lines in your code, especially:
您的代码中有一些不必要/错误的行，尤其是：
$query = mysql_query ("SELECT * FROM tb_funcionario WHERE
username='$username' and password='$password'");
$result = mysql_query($query);
I guess you wanted to write:
我猜你想写：
$query = "SELECT * FROM tb_funcionario WHERE
username='$username' and password='$password'";
$result = mysql_query($query);
回答by jere
your problem is here $query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
你的问题在这里 $query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
then you are running mysql_query($query)and it's trying to run the command against the resultset returned by the first statement, not a string like it should actually be
那么你正在运行mysql_query($query)，它试图针对第一个语句返回的结果集运行命令，而不是像它实际上应该的字符串
回答by Crontab
$query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
$result = mysql_query($query);
Is incorrect. You should just be assigning the query string to $query- what you're doing is running the query once, then trying to run the result of the query as a query again. You should have:
是不正确的。您应该只是将查询字符串分配给$query- 您所做的是运行一次查询，然后尝试再次将查询结果作为查询运行。你应该有：
$query = "SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'";
$result = mysql_query($query);
which would give you the results you seek.
这会给你你寻求的结果。
回答by Nobita
Look at this:
看这个：
$query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
$result = mysql_query($query);
Change it for:
将其更改为：
$query = "SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
$result = mysql_query($query);
回答by jbrtrnd
<?php
$query = "INSERT INTO tb_funcionario (nome_funcionario, username, password) VALUES (
'".mysql_real_escape_string($_POST[nome_funcionario])."',
'".mysql_real_escape_string($_POST[username])."',
'".mysql_real_escape_string($_POST[password])."'
)";
?>
回答by Yann Boisclair-Roy
Your problem comes from those lines:
你的问题来自这些行：
$query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");
$result = mysql_query($query);
Your calling the mysql_query()function twice.
您调用该mysql_query()函数两次。
You can do:
你可以做：
$sql = "SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'";
$result = mysql_query($sql);
Or either:
或者：
$result = mysql_query( "SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'"; );
