No results displayed when querying database
Question
I'm trying to create a searchable database using PHP and MySQL. I have a file called mission.html with the following code:
<html>
<body>
<form name="form1" method="post" action="mission1results.php" id="search">
<input name="search" type="text"/>
<input type="submit" name="submit" vaule="Search"/>
</form>
mission1results.php
<html>
<body>
<?php
include 'login.php';
$connection = mysqli_connect(
$db_hostname, $db_username,
$db_password, $db_database);
if(mysqli_connect_error()){
die("Database Connection Failed: " .
mysqli_connect_error() .
" (" . mysqli_connect_errno() . ")"
);
}
?>
<?php
$q_cond = mysqli_real_escape_string($_GET['search']);
$query="SELECT * From Merchant Where MerchantName='".$q_cond."'";
$result=mysqli_query($connection,$query);
if ($result===false)
{
die("Database Query Failed!")
};
while ($row=mysqli_fetch_assoc($result)){
echo "MerchantName: ".$row["MerchantName"].",";
echo "<hr/>";
}
mysqli_free_result($result);
?>
<?php
mysqli_close($connection);
?>
</body>
</html>
When I hit submit and type in anything in the searchbar nothing appears. I don't get an error, I don't get results, its all blank. Can anyone tell me why this is?
4 answers
solution1
0 2014-11-30 22:11:35
You have a syntax error in mission1results.php
if ($result===false)
{
die("Database Query Failed!")
};
must be changed for:
if ($result===false)
{
die("Database Query Failed!");
}
solution2
0 2014-11-30 22:21:08
First and foremost: mysqli_real_escape_string() requires a DB connection be passed, then there is your form where you are using a POST method in the form and GET for your query.
Consult the manual: http://php.net/manual/en/mysqli.real-escape-string.php
$q_cond = mysqli_real_escape_string($connection,$_POST['search']);
Plus, change
if ($result===false)
{
die("Database Query Failed!")
};
to
if ($result===false)
{
die("Database Query Failed!");
}
You also have a syntax error vaule="Search" change it to value
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Error reporting should only be done in staging, and never production.
Also or die(mysqli_error($connection)) to mysqli_query() to find any possible errors.
solution3
0 2014-11-30 22:25:03
Instead
$_GET['search']use$_POST['search']because your submit forms method ispost.One of
mysqli_real_escape_stringparameters should be DB connection.syntax errors in HTML, for example,
vaule="Search"syntax errors in PHP, for example, there shoudn't be
;after}inif
solution4
0 2014-11-30 22:43:03
If you are getting a blank screen with the errors pointed out in previous answers you might want to take a look at the PHP error_reporting level on your system http://php.net/manual/en/function.error-reporting.php . You should be seeing PHP errors, on a development server I like to report PHP errors, warnings and notices.
Also, are you expecting users to enter an exact search term? You might want to consider something like:
$query="SELECT * From `Merchant` Where `MerchantName` like '%".$q_cond."%'";
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