I am having problems trying to get these queries with a WHERE clause to work. I have two tables which look like this :
What I am trying to do is return the genre that each film has. At the moment no data is returning at all from what I can see. Here are the two queries:
$film_id = $row_movie_list['film_id'];
mysql_select_db($database_fot , $fot);
$query_get_genre = "SELECT * FROM film_genre WHERE `id_film` ='". $film_id. "'";
$get_genre = mysql_query($query_get_genre, $fot) or die(mysql_error());
$row_get_genre = mysql_fetch_assoc($get_genre);
$totalRows_get_genre = mysql_num_rows($get_genre);
$genre_id = $row_get_genre['id_genre'];
mysql_select_db($database_fot , $fot);
$query_genre = "SELECT * FROM genre WHERE `id_genre` ='". $genre_id. "'";
$genre= mysql_query($query_genre, $fot) or die(mysql_error());
$row__genre = mysql_fetch_assoc($genre);
$totalRows_genre = mysql_num_rows($genre);
PHP with content area. I fairly new to PHP so any help would be appreciated.
<?php do { echo $genre['genre']; } while($row_get_genre = mysql_fetch_assoc($get_genre)); ?>
Update: I am now able to get first genre but not second it just echos the first one twice and I have tried but still no luck:
do {do { echo $row_genre['genre']; } while($row_genre = mysql_fetch_assoc($genre));} while($row_get_genre = mysql_fetch_assoc($get_genre)); ?>
Avoiding the fact that you're using a deprecated way to establish connection and interact with MySQL, what you're doing is getting a single relation genre-film and then getting the row of the genre that matches. You should surround part of your code with a while that executes while it's still genres of the film with id. Something like:
$film_id = $row_movie_list['film_id'];
mysql_select_db($database_fot , $fot);
$query_get_genre = "SELECT * FROM film_genre WHERE `id_film` ='". $film_id. "'";
$get_genre = mysql_query($query_get_genre, $fot) or die(mysql_error());
while($row_get_genre = mysql_fetch_assoc($get_genre)){
$genre_id = $row_get_genre['id_genre'];
$query_genre = "SELECT * FROM genre WHERE `id_genre` ='". $genre_id. "'";
$genre= mysql_query($query_genre, $fot) or die(mysql_error());
$row__genre = mysql_fetch_assoc($genre);
// You should do whatever you want to do with $row__genre here. Otherwise it will be cleared.
}
I must insist this is a deprecated and insecure way of communication with a MySQL Database. I recommend you read about MySQLi or PDO extensions.
MySQLi: http://www.php.net/manual/en/book.mysqli.php
PDO: http://www.php.net/manual/en/book.pdo.php
I began to create a website for my small real estate business.
I played a bit with functions http://www.php.net mysql and I managed to make a page accessed via AJAX and returning html content for the search engine.
I have a database already populated with apartments and houses
The problem is that if the apartment name is "apartment" I return html content if "apartment with 3 rooms" it no longer write anything.
I do not understand where I was wrong:
<?php
$search = $_GET['selected'];
$link = mysql_connect('localhost', 'root', '');
mysql_select_db('houses', $link);
function searchHouse($search, $link){
$query = "select * from houses where name=$search limit 1";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
$query2 = "select * from houses_info where house_id=$row[id]";
$result2 = mysql_query($query2);
$row = mysql_fetch_assoc($result2);
return $row;
}
$result = searchHouse($search, $link);
echo $result['house_sq'];
echo "<br>";
echo $result['house_rooms'];
echo "<br>";
echo $result['house_bathrooms'];
echo "<br>";
echo $result['house_address'];
?>
you should know if you "played" with php.net that mysql_* functions are deprecated and are no longer maintained. It's a red box on top of the page informing you that.
you have a big MySQL injection hole there, you are not escaping $string at all
your problem is that you are not adding quotes to $string like: '$string'
you should stat using PDO to get rid of the bad code and SQL Injections holes.
you can wrap those 2 selects into a single select:
<?php
function searchHouse($search, $link){
$search = mysql_real_escape_string($search);
$query = "select * from houses_info where house_id IN (select * from houses where name='".$search."' limit 1)";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
return $row;
}
?>
since you are already building that website you can start moving to PDO, read this tutorial, your code will be more like this:
<?php
$db = new PDO('mysql:host=localhost;dbname=houses;charset=UTF-8', 'root', '', array(PDO::ATTR_EMULATE_PREPARES => false, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));
$search = $_GET['selected'];
function searchHouse($search){
global $db;
$query = $db->prepare("select * from houses_info where house_id IN (select * from houses where name=:search limit 1)");
$query->execute(array(':search' => $search));
$row = $query->fetch(PDO::FETCH_ASSOC);
return $row;
}
$result = searchHouse($search);
?>
try:
$query = "select * from houses where name='".mysql_real_escape_string($search)."' limit 1";
and remember to always sanitize user input before passing it to sql to avoid sql injections.
Your first query should be:
$query = "select * from houses where name like $search% limit 1";
Strings need to be quoted in queries. Also, this is vulnerable to MySQL injection, make sure to escape $search with mysql_real_escape_string. Or even better yet use MySQLi or PDO instead of the old mysql_ functions.
$query = "select * from houses where name=$search limit 1";
Should be:
$query = "select * from houses where name='$search' limit 1";
Although you REALLY need to escape $search because it came from a user, even if they aren't malicious, any search queries with a single quote in it will break;
$search = $_GET['selected'];
Should be:
$search = mysql_real_escape_string($_GET['selected']);
(Anybody have the copy paste handy with the links to tutorials for MySQLi/PDO and such?)
I'm having slight difficulties (syntax presumably) formulating a query for a search I'm writing in php.
So far I have this:
$query = ("SELECT * FROM $table WHERE $field LIKE "$trimmed);
trimmed is defined as
$trimmed = trim($var);
What I'm trying to accomplish is, use that query to search for a certain row in my mysql database. I've confirmed that it does indeed connect to the dbase and it does grab data from the table. I'm 99% new to php and mysql, I've just started working on this. Any help would be greatly appreciated.
EDIT: Oh I use the query here:
$result = mysql_query($query); I'm sure the issue isn't here, but in $query
Change
$query = ("SELECT * FROM $table WHERE $field LIKE "$trimmed);
to
$query = "SELECT * FROM $table WHERE $field LIKE '$trimmed'";
It's always a good idea to escape any special characters, such as backslash, in the input. With mysql, you can use mysql_escape_string:
$trimmed = mysql_escape_string($trimmed);
$query = "SELECT * FROM $table WHERE $field LIKE '$trimmed'";
Equivalent commands exist in mysqli, PDO, and all PHP frameworks.
Check out the PHP manul, example:
$query= mysql_query("SELECT data FROM mydb;");
$myarray= array();
while ($row= mysql_fetch_array($query)) {
$myarray[] = $row['data'];
}
EDIT
This is your code? if so, you have a syntax error:
$query = ("SELECT * FROM $table WHERE $field LIKE "$trimmed);
should be:
$query = ("SELECT * FROM $table WHERE $field LIKE '$trimmed'");
What's the best way with PHP to read a single record from a MySQL database? E.g.:
SELECT id FROM games
I was trying to find an answer in the old questions, but had no luck.
This post is marked obsolete because the content is out of date. It is not currently accepting new interactions.
$id = mysql_result(mysql_query("SELECT id FROM games LIMIT 1"),0);
$link = mysql_connect('localhost','root','yourPassword')
mysql_select_db('database_name', $link);
$sql = 'SELECT id FROM games LIMIT 1';
$result = mysql_query($sql, $link) or die(mysql_error());
$row = mysql_fetch_assoc($result);
print_r($row);
There were few things missing in ChrisAD answer. After connecting to mysql it's crucial to select database and also die() statement allows you to see errors if they occur.
Be carefull it works only if you have 1 record in the database, because otherwise you need to add WHERE id=xx or something similar to get only one row and not more. Also you can access your id like $row['id']
Using PDO you could do something like this:
$db = new PDO('mysql:host=hostname;dbname=dbname', 'username', 'password');
$stmt = $db->query('select id from games where ...');
$id = $stmt->fetchColumn(0);
if ($id !== false) {
echo $id;
}
You obviously should also check whether PDO::query() executes the query OK (either by checking the result or telling PDO to throw exceptions instead)
Assuming you are using an auto-incrementing primary key, which is the normal way to do things, then you can access the key value of the last row you put into the database with:
$userID = mysqli_insert_id($link);
otherwise, you'll have to know more specifics about the row you are trying to find, such as email address. Without knowing your table structure, we can't be more specific.
Either way, to limit your SELECT query, use a WHERE statement like this:
(Generic Example)
$getID = mysqli_fetch_assoc(mysqli_query($link, "SELECT userID FROM users WHERE something = 'unique'"));
$userID = $getID['userID'];
(Specific example)
Or a more specific example:
$getID = mysqli_fetch_assoc(mysqli_query($link, "SELECT userID FROM users WHERE userID = 1"));
$userID = $getID['userID'];
Warning! Your SQL isn't a good idea, because it will select all rows (no WHERE clause assumes "WHERE 1"!) and clog your application if you have a large number of rows. (What's the point of selecting 1,000 rows when 1 will do?) So instead, when selecting only one row, make sure you specify the LIMIT clause:
$sql = "SELECT id FROM games LIMIT 1"; // Select ONLY one, instead of all
$result = $db->query($sql);
$row = $result->fetch_assoc();
echo 'Game ID: '.$row['id'];
This difference requires MySQL to select only the first matching record, so ordering the table is important or you ought to use a WHERE clause. However, it's a whole lot less memory and time to find that one record, than to get every record and output row number one.
One more answer for object oriented style. Found this solution for me:
$id = $dbh->query("SELECT id FROM mytable WHERE mycolumn = 'foo'")->fetch_object()->id;
gives back just one id. Verify that your design ensures you got the right one.
First you connect to your database. Then you build the query string. Then you launch the query and store the result, and finally you fetch what rows you want from the result by using one of the fetch methods.
$link = mysql_connect('localhost','root','yourPassword')
mysql_select_db('database',$link);
$sql = 'SELECT id FROM games'
$result = mysql_query($sql,$link);
$singleRow = mysql_fetch_array($result)
echo $singleRow;
Edit: So sorry, forgot the database connection. Added it now
'Best way' aside some usual ways of retrieving a single record from the database with PHP go like that:
with mysqli
$sql = "SELECT id, name, producer FROM games WHERE user_id = 1";
$result = $db->query($sql);
$row = $result->fetch_row();
with Zend Framework
//Inside the table class
$select = $this->select()->where('user_id = ?', 1);
$row = $this->fetchRow($select);
The easiest way is to use mysql_result.
I copied some of the code below from other answers to save time.
$link = mysql_connect('localhost','root','yourPassword')
mysql_select_db('database',$link);
$sql = 'SELECT id FROM games'
$result = mysql_query($sql,$link);
$num_rows = mysql_num_rows($result);
// i is the row number and will be 0 through $num_rows-1
for ($i = 0; $i < $num_rows; $i++) {
$value = mysql_result($result, i, 'id');
echo 'Row ', i, ': ', $value, "\n";
}
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$db = new mysqli('localhost', 'tmp', 'tmp', 'your_db');
$db->set_charset('utf8mb4');
if($row = $db->query("SELECT id FROM games LIMIT 1")->fetch_row()) { //NULL or array
$id = $row[0];
}
I agree that mysql_result is the easy way to retrieve contents of one cell from a MySQL result set. Tiny code:
$r = mysql_query('SELECT id FROM table') or die(mysql_error());
if (mysql_num_rows($r) > 0) {
echo mysql_result($r); // will output first ID
echo mysql_result($r, 1); // will ouput second ID
}
Easy way to Fetch Single Record from MySQL Database by using PHP List
The SQL Query is SELECT user_name from user_table WHERE user_id = 6
The PHP Code for the above Query is
$sql_select = "";
$sql_select .= "SELECT ";
$sql_select .= " user_name ";
$sql_select .= "FROM user_table ";
$sql_select .= "WHERE user_id = 6" ;
$rs_id = mysql_query($sql_select, $link) or die(mysql_error());
list($userName) = mysql_fetch_row($rs_id);
Note: The List Concept should be applicable for Single Row Fetching not for Multiple Rows
Better if SQL will be optimized with addion of LIMIT 1 in the end:
$query = "select id from games LIMIT 1";
SO ANSWER IS (works on php 5.6.3):
If you want to get first item of first row(even if it is not ID column):
queryExec($query) -> fetch_array()[0];
If you want to get first row(single item from DB)
queryExec($query) -> fetch_assoc();
If you want to some exact column from first row
queryExec($query) -> fetch_assoc()['columnName'];
or need to fix query and use first written way :)
im doing some queries in Zend Framework and i need to make sure no SQL injection is possible in the next kind of formats. I can use mysql_escape(deprecated) and wont do all the work. If i try to use real_mysql_escape it wont be able to grab the conection with the database and i cant find how zend_filter would solve the problem.
The query im doing (simplied) have the next sintaxes:
$db = Zend_Registry::get('db');
$select = "SELECT COUNT(*) AS num
FROM message m
WHERE m.message LIKE '".$username." %'";
$row = $db->fetchRow($select);
What is the best way to prevent SQL INJECTION with this framework?
Easy:
$db->quote($username);
So:
$username = $db->quote($username . '%');
$select = 'SELECT COUNT(*) AS num
FROM message m
WHERE m.message LIKE ' . $username;
$row = $db->fetchRow($select);
$sql = 'SELECT * FROM messages WHERE username LIKE ?';
$row = $db->fetchRow($sql, $username);
Reference: http://framework.zend.com/manual/en/zend.db.html
When working with a model you can use:
$bugs = new Bugs();
$row = $bugs->fetchRow($bugs->select()->where('bug_id = ?', 1));