MYSQL WHERE clause is wrong value - php

Whenever I try to perform my query, It gives me an unknown column error, because it is using my variable as the column name.
essentially
$search="lname";
$term="asdas";
(both of those are variables from a form on another page)
I run this:
if (isset($term))
{
$query = "SELECT * FROM test
WHERE $search = $term ";
}
else
{
$query = "SELECT * FROM test";
}
echo $query;
$result=mysql_query($query) or die(mysql_error());
and then I get this as my error:
Unknown column 'asdas' in 'where clause'

You need to enclose the search term in single quotes(also use mysql_real_escape_string to avoid any issues with quotes in the search string.).
i.e:
if (isset($term))
{
$query = "SELECT * FROM test WHERE $search = '" . mysql_real_escape_string($term) . "' ";
}

You need to quote it.
if (isset($term))
{
$query = "SELECT * FROM test
WHERE $search = '$term' ";
}
else
{
$query = "SELECT * FROM test";
}
echo $query;
$result=mysql_query($query) or die(mysql_error());
Other comments
It is always better to use parameterized queries if the driver supports it. It will prevent SQL injection. As it stands, someone could send in a string "' or ''='" and the query turns out to be
SELECT * FROM test WHERE col1 = '' or ''=''
which is really benign but unexpected behaviour. If the string contains single quotes, it also breaks your query (input is "o'neil")
SELECT * FROM test WHERE col1 = 'o'neil' # << unmatched quotes
So, at the very least use mysql_real_escape_string if you cannot use parameters, i.e.
$query = "SELECT * FROM test
WHERE $search = '" . mysql_real_escape_string($term) . "' ";

You need to quote your $term parameter:
// protect from trivial sql injection attacks.
$term = mysql_real_escape_string("adas");
$query = "SELECT * FROM test
WHERE $search = '$term'";

You have to surround the term value with quotes:
SELECT *
FROM test
WHERE lname='asdas'
otherwise any SQL server out there will think asdas is a field name and try to find it in the table.

Add ' around your columns
$query = "SELECT * FROM test WHERE $search = '$term' ";

you need to put single quotes around $term so that the SQL thinks it's a string

put single quote string always be quoted. Do not forgot use mysql_real_escape_sring()
$query = "SELECT * FROM test
WHERE $search = '$term' ";

Put single quotes around $term
if (isset($term))
{
$query = "SELECT * FROM test WHERE $search = '$term'";
}
else
{
$query = "SELECT * FROM test";
}
echo $query;
$result=mysql_query($query) or die(mysql_error());

Related

How to Concatenate table name with with a variable value in mySQL

I'm trying to create a dynamic code that would ready from any table with the certain name but the difference between each table name is a number that is generated by a variable: for example :
//that's how I get my variable the value for example is = 3
$pid = $GLOBALS["localid"];
//the table name for example is tablename_3
$strTable = "tablename_" .$pid;
//here's how the query should look like
$query = "SELECT * FROM . $strTable . where .....;
I'm making a mistake somewhere but can't figure it out and would appreciate a little help please
Remove the dots and also make sure you have single quotes aroung where
$query = "SELECT * FROM $strTable where '.....';
Besides the comments about do or don't build your queries like this...
You're not closing the quotes properly.
$query = "SELECT * FROM . $strTable . where .....; //Double quote not closed.
should be:
$query = 'SELECT * FROM' . $strTable . 'where .....'; //Single quoted strings concatenated with variable.
or
$query = "SELECT * FROM $strTable where ....."; //Variable inside double quoted string.

Error undeclared variable when select using limit

I have an error in my script select MySQL with limit. it said
Gagal ambil data:Undeclared variable: $st
this is my script :
include'../konekdb.php';
if(empty($_GET[start]))
{
$st="0";
}
else
{
$st=$_GET[start];
}
$query = 'SELECT * FROM pelanggan LIMIT $st,5';
$ambildata = mysql_query($query);
if(!$ambildata)
{
die('Gagal ambil data:'.mysql_error());
}
include'tabel_pelanggan.php';
mysql_free_result($ambildata);
mysql_close($koneksi);
$query2 = mysql_query("SELECT * FROM pelanggan");
$num = mysql_num_rows($query2);
$hal = ceil($num/5);
echo "Halaman :";
for($i=1;$i<=$hal;$i++){
$page=$i-1;
echo "[][<a href=$_SERVER[PHP_SELF]?start=$page>$i</a>][]";
}
can you help me? thanks before :)
You have an error with you $_GET, you're missing the quotes around the key name.
Your $_GET,
$st=$_GET[start];
What it should be,
$st = $_GET['start'];
Edit 1
You shouldn't be using MySQL now as it is deprecated. Either start using MySQLi or PDO, you should also look in to prepared statements.
Edit 2
Also don't forget to change your if statement too.
From
if(empty($_GET[start]))
To,
if(empty($_GET['start']))
Edit 3
Your query needs to use double quotes or if you want to use single quotes you need to close the single quotes then add the variable on.
From,
$query = 'SELECT * FROM pelanggan LIMIT $st,5';
To,
$query = 'SELECT * FROM pelanggan LIMIT ' . $st . ',5';
Or,
$query = "SELECT * FROM pelanggan LIMIT {$st},5";

How to run an sql query inside a while loop of another query

The following works with no problem at all, when the photoId is directly on the statement and not a variable.
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "103"') or die(mysqli_error($con));
but the following just won't work with no error, what might be causing this not to select.
$imageid = '103';
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
echo $img_row['img'];
This is inside a while loop.
while($row = mysqli_fetch_array($somequery)){
$imageid = $row['photoid'];
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
echo $img_row['img'];
}
Thanks.
in php a ' and a " are very different and the query syntax is double quote around the query and single quote around variables.. although I would recommend you look at using parameters on your query instead of just putting a variable directly into the query
Per my recommendation you should change your query to this:
$imageid = '103';
$query = $con->prepare("SELECT * FROM imgs WHERE photoid = ?");
$query->bind_param('sssd', $imageid);
$query->execute();
this is just the nuts and bolts of it... if you want more information about the connection.. error handling and everything else read the DOCS
there is a big difference between ' and " in php
Differences
change your query to be
$img_query = mysqli_query($con, "SELECT * FROM imgs WHERE photoid = '$imageid'") or die(mysqli_error($con));
and it should work.

PHP not displaying results from MySQL database

I am trying to display an entry from a MySql database which is selected by GET data.
if (isset($_GET["id"])){
$id=$_GET["id"];
$result = getSelectedBlog($id);
while($row = mysqli_fetch_array($result))
{
extract($row);
?>
<div class="headline"><?php echo $headline ?></div>
<div class="subtitle"><?php echo $subTitle ?></div>
<div class="content"><?php echo $content ?></div>
<?php
}
Here is the SQL statement:
function getSelectedBlog($id){
$con = mysqli_connect('localhost', 'root', '', 'michaelWebsite') or die('could not connect');
$sql = 'SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE "$id"';
$result = mysqli_query($con, $sql) or die('entry does not exist.:' . mysqli_error($con));
return $result;
}
As you can see, I am passing the get data as $id to the method that returns the result. However nothing is being returned. There are three entries at the moment, if I change $id in the SQL statement to either 1, 2 or 3 it will show the corresponding data but it just will not work with the $id variable.
The URL does end with the correct info ?id=1.
Please excuse me if it is something stupid, I have just been stuck on this for hours now!!
All of these answers will solve your problem, but none have mentioned or prevented SQL Injection.
In your case I recommend (assuming articleID is an integer field).
$sql = 'SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE "' . (int)$id . '"';
I'm also curious why you are using LIKE for an id field.
Note: Since you are using MySQLi, I'd encourage you to look at prepared statements.
$sql = 'SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE "'.$id.'"';
escape your var in simple quote
Try with this:
$sql = "SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE '$id'";
or with
$sql = 'SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE "' . $id . '"';
You need to use double quotes in order for php to correctly expand your variables :) so change your query to
$sql = "SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE '$id'";
Change
'SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE "$id"'
to
"SELECT * FROM tblArticle WHERE tblArticle.articleID LIKE '$id'"
Variables will be evaluated only if they're between double quotes "

php with SQL query

i have a problem with php in the following:
$sql = 'SELECT name FROM chiled WHERE `im` LIKE $id ';
$query = mysql_query( $sql );
$a=mysql_fetch_row($query);
echo $a[0];
there is error in mysql_fetch_row($query);
but if i do the following :
$sql = 'SELECT name FROM chiled WHERE `im` LIKE 1111 ';
$query = mysql_query( $sql );
$a=mysql_fetch_row($query);
echo $a[0];
it is working and prints the name
can you please tell me what is wrong?
Single quotes in PHP doesn't evaluate embedded variables - you need to use double quotes to do that. (See the "Single quoted" section of the PHP Strings manual page for more info..)
i.e.: $sql = "SELECT name FROM chiled WHERE 'im' LIKE $id ";
Or better still...
$sql = 'SELECT name FROM chiled WHERE im="' . mysql_real_escape_string($id) . '"';
(As you're not using the % in your like, you're presumably not attempting to do any form of pattern matching.)
Additionally, I'd recommend a read of the existing Best way to stop SQL Injection in PHP question/answers.
Are you sure you want to be using LIKE? It looks more to me like you want to see if im = $id. Also, make sure you're escaping your variables before using them in the query.
Edit
If you DO want to us LIKE, you probably want something like this:
$sql = "SELECT name FROM chiled WHERE `im` LIKE '%$id%' ";
which will find anywhere that the string $id is found in the im column.
You need to quote the variable after LIKE, like this:
$sql = "SELECT name FROM chiled WHERE im LIKE '$id'";
$query = mysql_query($sql);
$a = mysql_fetch_row($query);
echo $a[0];
// ....
Beside, you are using single quotes, Therefore, $id is not replaced for its value. Your query look like this:
SELECT name FROM chiled WHERE im LIKE $id;
$sql = "SELECT name FROM chiled WHERE `im` LIKE '$id' ";
change to double quotes - http://php.net/manual/en/language.types.string.php

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