问题
I try to check if the username and email from input fields already exists in my database before to create a new user.
this is login.php :
<?php
session_start();
$pseudo = $_POST['pseudo'];
$mail = $_POST['mail'];
$pseudo=mysql_real_escape_string($pseudo);
$pseudo=ltrim($pseudo);
$pseudo=rtrim($pseudo);
$mail=mysql_real_escape_string($mail);
$mail=trim($mail);
$sql=mysqli_connect('localhost','root','','bdd_name');
$query=("SELECT COUNT(*) FROM t_people WHERE 'PEO_PSEUDO'='".$pseudo."' OR 'PEO_MAIL'='".$mail."'");
$result = mysqli_prepare($sql,$query);
mysqli_stmt_execute($result);
mysqli_stmt_store_result($result);
if (mysqli_stmt_num_rows($result) == 0) {
echo 1;
}
else {
echo 2;
}
mysqli_stmt_free_result($result);
mysqli_stmt_close($result);
?>
And this is a part of my JavaScript :
var pseudo=$("#pseudo").val();
var mail=$("#mail").val();
$.ajax({
type: "POST",
url: "login.php",
data: {pseudo:pseudo, mail:mail}
}).done(function(result) {
if (result==1) {
good();
}
else if (result==2) {
bad();
}
});
Can anyone tell me what is wrong with this?
I'm on this since hours now and I'm clueless...
回答1:
There are some things going wrong.
Don't use mysql_real_escape_string
cause you're working with mysqli_*
. Use mysqli_real_escape_string
instead. But better use mysqli_stmt_bind_param
because you're working with prepared statements. And if you work with COUNT(*)
you always get 1 row.
$pseudo = $_POST['pseudo'];
$mail = $_POST['mail'];
$query = "SELECT * FROM t_people WHERE PEO_PSEUDO = ? OR PEO_MAIL = ? LIMIT 1";
$stmt = mysqli_prepare($sql, $query);
mysqli_stmt_bind_param($stmt, 'ss', $pseudo, $mail);
mysqli_stmt_execute($stmt);
mysqli_stmt_store_result($stmt);
$numRows = mysqli_stmt_num_rows($stmt);
mysqli_stmt_close($stmt);
With COUNT(*)
(which is more efficient) it goes like:
$query = "SELECT COUNT(*) as numrows FROM t_people WHERE PEO_PSEUDO = ? OR PEO_MAIL = ?";
...
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $numRows);
mysqli_stmt_fetch($stmt);
mysqli_stmt_close($stmt);
// work with $numRows
回答2:
You are using wrong operator for escaping use backtick operator in your query like this
"SELECT COUNT(*) FROM t_people WHERE `PEO_PSEUDO`='".$pseudo."' OR `PEO_MAIL`='".$mail."'"
来源:https://stackoverflow.com/questions/16384345/check-if-the-username-already-exists-with-mysqli