PHP MySQL Query Where x = $variable [closed]

问题

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Closed 6 years ago.

I have this code (I know that the email is defined)

 <?php
$con=mysqli_connect($host,$user,$pass,$database);
 if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '.$email.'");

while($row = mysqli_fetch_array($result))
echo $row
?>

In my MySQL database I have the following setup (Table name is glogin_users) id email note

I've tried extracting the note text from the database and then echo'ing it but it doesn't seem to echo anything.

回答1:

What you are doing right now is you are adding . on the string and not concatenating. It should be,

$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '".$email."'");

or simply

$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '$email'");

回答2:

You have to do this to echo it:

echo $row['note'];

(The data is coming as an array)

回答3:

$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '".$email."'");
while($row = mysqli_fetch_array($result))
echo $row['note'];

来源：https://stackoverflow.com/questions/15703608/php-mysql-query-where-x-variable