I want to take the column names from a table and display them, so i can compare them later.
To get the names, i tried:
$entry = mysqli_query($con, 'SHOW COLUMNS FROM table');
and
$entry = mysqli_query($con, "SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_SCHEMA = '$db' AND TABLE_NAME = 'table'");
I don't know whether this runs correctly or not, since i don't get an error message there.
If i try to print the contents of $entry via echo, i keep getting errors.
Previously in my code, i print other entries using:
$test = mysqli_query($con, 'SELECT DISTINCT LK_Release FROM table');
while($row = mysqli_fetch_object($test))
{
echo "Releasename: " . "$row->LK_Release". "<br>";
... }
This output works for me.
What i tried to output the columnnames:
while($row = mysqli_fetch_object($entry))
{
echo $row;
}
Any ideas?
Your query is already correct. But you lack something in the fetching:
while($row = mysqli_fetch_object($entry))
{
echo $row->Field . '<br/>';
// ^ access the objects properties
}
The following SQL statements are nearly equivalent:
SELECT COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_name = 'tbl_name'
You can use DESCRIBE:
DESCRIBE my_table;
Or in newer versions you can use INFORMATION_SCHEMA:
SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA = 'my_database' AND TABLE_NAME = 'my_table';
Or you can use SHOW COLUMNS:
SHOW COLUMNS FROM my_table;
I am kinda new to PHP and MySql and I am trying to find a way to echo the table names of my database in my page.
When I use : SELECT table_name FROM INFORMATION_SCHEMA.TABLES WHERE table_schema = 'myDBname'
directly in PHPMyAdmin, I get the results I want but I just don't know I to "echo" it in my page.
This is what I am using at the moment :
$statement = $db->prepare("SELECT table_name FROM INFORMATION_SCHEMA.TABLES WHERE table_schema = 'myDBname'");
$row = $statement->fetchAll();
What would be, in your opinion, the best way to display it on my page if I want to eventually echo the result of the query in a dropdown menu ?
This works for me.
$tables = array();
$stmt = $db->query("SHOW TABLES");
while($row = $stmt->fetch(PDO::FETCH_NUM)){
$tables[] = $row[0];
}
var_dump($tables);
I'm trying to display a field from my MySQL database. It's in the table tblproducts in the row with the id is set to 1. And under the column qty.
This is the code I'm using:
<?php
mysql_connect("localhost","username","password");
mysql_select_db("database_name");
$available = "SELECT qty FROM tblproducts WHERE id = 1";
$result = mysql_query($available);
echo $result;
?>
However, I keep getting this message: Resource id #2
I've done a bit of research and seen where other people are having similar problems but most of them are trying to display their data in an HTML table whereas I just need the data from 'qty' to display. And of course I'm definitely not a MySQL guru.
Can anyone help me out with this please?
Try changing this:
$result = mysql_query($available);
To this:
$result = mysql_result(mysql_query($available), 0);
Let's start from the start. (I'll assume you have the connection set)
Form the query
$query = "SELECT `qty`
FROM `tblproducts`
WHERE `id` = 1";
Execute the query
$run = mysql_query($query);
Now, put the result in an assoc array
$r = mysql_fetch_array($run);
See the contents of the array
echo $r['qty'];
It's also advised that you move up from mysql to either mysqli, or PDO. PDO is preferred as you're not bound to the MySQL database model.
Try this:
Here you need to generate associative array and then get the resulting row.
$query = "SELECT `qty` FROM `tblproducts` WHERE `id` = 1";
$run = mysql_query($query);
$r = mysql_fetch_array($run);
echo $r['qty'];
-
Thanks
What's the best way with PHP to read a single record from a MySQL database? E.g.:
SELECT id FROM games
I was trying to find an answer in the old questions, but had no luck.
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$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 :)
What is the best way to check if a table exists in MySQL (preferably via PDO in PHP) without throwing an exception. I do not feel like parsing the results of "SHOW TABLES LIKE" et cetera. There must be some sort of boolean query?
Querying the information_schema database using prepared statement looks like the most reliable and secure solution.
$sql = "SELECT 1 FROM information_schema.tables
WHERE table_schema = database() AND table_name = ?";
$stmt = $pdo->prepare($sql);
$stmt->execute([$tableName]);
$exists = (bool)$stmt->fetchColumn();
If you're using MySQL 5.0 and later, you could try:
SELECT COUNT(*)
FROM information_schema.tables
WHERE table_schema = '[database name]'
AND table_name = '[table name]';
Any results indicate the table exists.
From: http://www.electrictoolbox.com/check-if-mysql-table-exists/
Using mysqli I've created following function. Assuming you have an mysqli instance called $con.
function table_exist($con, $table){
$table = $con->real_escape_string($table);
$sql = "show tables like '".$table."'";
$res = $con->query($sql);
return ($res->num_rows > 0);
}
Hope it helps.
Warning: as sugested by #jcaron this function could be vulnerable to sqlinjection attacs, so make sure your $table var is clean or even better use parameterised queries.
This is posted simply if anyone comes looking for this question. Even though its been answered a bit. Some of the replies make it more complex than it needed to be.
For mysql* I used :
if (mysqli_num_rows(
mysqli_query(
$con,"SHOW TABLES LIKE '" . $table . "'")
) > 0
or die ("No table set")
){
In PDO I used:
if ($con->query(
"SHOW TABLES LIKE '" . $table . "'"
)->rowCount() > 0
or die("No table set")
){
With this I just push the else condition into or. And for my needs I only simply need die. Though you can set or to other things. Some might prefer the if/ else if/else. Which is then to remove or and then supply if/else if/else.
Here is the my solution that I prefer when using stored procedures. Custom mysql function for check the table exists in current database.
delimiter $$
CREATE FUNCTION TABLE_EXISTS(_table_name VARCHAR(45))
RETURNS BOOLEAN
DETERMINISTIC READS SQL DATA
BEGIN
DECLARE _exists TINYINT(1) DEFAULT 0;
SELECT COUNT(*) INTO _exists
FROM information_schema.tables
WHERE table_schema = DATABASE()
AND table_name = _table_name;
RETURN _exists;
END$$
SELECT TABLE_EXISTS('you_table_name') as _exists
As a "Show tables" might be slow on larger databases, I recommend using "DESCRIBE " and check if you get true/false as a result
$tableExists = mysqli_query("DESCRIBE `myTable`");
$q = "SHOW TABLES";
$res = mysql_query($q, $con);
if ($res)
while ( $row = mysql_fetch_array($res, MYSQL_ASSOC) )
{
foreach( $row as $key => $value )
{
if ( $value = BTABLE ) // BTABLE IS A DEFINED NAME OF TABLE
echo "exist";
else
echo "not exist";
}
}
Zend framework
public function verifyTablesExists($tablesName)
{
$db = $this->getDefaultAdapter();
$config_db = $db->getConfig();
$sql = "SELECT COUNT(*) FROM information_schema.tables WHERE table_schema = '{$config_db['dbname']}' AND table_name = '{$tablesName}'";
$result = $db->fetchRow($sql);
return $result;
}
If the reason for wanting to do this is is conditional table creation, then 'CREATE TABLE IF NOT EXISTS' seems ideal for the job. Until I discovered this, I used the 'DESCRIBE' method above. More info here: MySQL "CREATE TABLE IF NOT EXISTS" -> Error 1050
Why you make it so hard to understand?
function table_exist($table){
$pTableExist = mysql_query("show tables like '".$table."'");
if ($rTableExist = mysql_fetch_array($pTableExist)) {
return "Yes";
}else{
return "No";
}
}