Created a new variable by pulling a value from a MySQL table - php

I am using a page where a variable $submissionid is being posted to it. I would like to use this variable and pull the field subcheck from a MySQL table called submission. I would like the value of the field subcheck to simply be a new variable $subcheck. I'm not sure how to do this. I have a query below, but how to I convert the result of the query below into a variable called $subcheck? (For any given submissionid, there will only be one $subcheck.)
Thanks in advance,
John
$querysub = mysql_query("SELECT subcheck FROM submission WHERE submissionid = '$submissionid' ");
mysql_query($querysub) or die(mysql_error());

You can try:
$querysub = mysql_query("SELECT subcheck FROM submission WHERE submissionid = ".
mysql_real_escape_string($submissionid));
$result = mysql_query($querysub);
if (!$result) {
die 'Could not run query: ' . mysql_error();
}
$subcheck = mysql_result($result, 0);

This is more of a 'php' question, than it is for mysql.
Look up the 'extract' keyword for PHP Link. Effectively 'extract' takes the contents of an associative array and creates php variables (symbol table entries) using the names of keys. Each php variable will then contain the associated value.
You should be able to just:
$result = mysql_query("SELECT * FROM table");
$row = mysql_fetch_array( $result, MYSQL_ASSOC );
extract( $row ); // Create php variables, named after each column in the table.
$row["field"] == $field; // Will be a true statement after 'extract()'
Enjoy, you now have the ability to have your code dynamic adjust to a DB schema that could be changed.
-- J Jorgenson --

This should work:
$querysub = mysql_query("SELECT subcheck FROM submission WHERE submissionid = '" . $submissionid ."' ");
$result = mysql_query($querysub) or die(mysql_error());
$row = mysql_fetch_assoc( $result );
if ($row ) {
$subcheck = $row['subcheck'];
} else {
echo "Subcheck not found";
}
Be careful with the escape characters around $submissionid in your query string. In your sample, they are probably letting the name of the variable go into the string you send to the mysql server.

Related

mySql query does not find the row

I'm new to both PHP & mySQL, but I suspect some apostrophe related bug here (maybe).
For some reason the first query always seems to return null because the
echo "result: $result\n";
never prints any data. At the first call that is expected, but at the second call the player has been added to the db. My database works in the sense that I can see that rows with correct ids are added to the database (via phpMyAdmin).
Can you help me spot the error?
<?php
require_once('Db.php');
$db = new Db();
// Quote and escape form submitted values
$id = $db->quote($_POST['id']);
$score = $db->quote($_POST['score']);
$result = $db->query("SELECT `id` FROM `xxxxxx`.`Player` WHERE `id` = `$id`");
echo "result: $result\n"; // Never prints any data
if($result->num_rows == 0) {
// Row not found. Create it!
$result = $db->query("INSERT INTO `xxxxxx`.`Player` (`id`,`score`) VALUES (" . $id . "," . 0 . ")");
}
?>
First, drop those backticks from id in WHERE clause, otherwise it will take the field name from id column instead of 'id'.
Then you need to fetch data from $result:
$result = $db->query("SELECT id FROM `xxxxxx`.`Player` WHERE id = '$id'");
$row = $result->fetch_array();
echo $row['id'];
Or if there are more rows than one:
while($row = $result->fetch_array())
{
echo $row['id'];
}
You are using backticks in your query for $id. Remove them and try again.
Your query should be
$result = $db->query("SELECT `id` FROM `xxxxxx`.`Player` WHERE `id` = $id");
OR
$result = $db->query("SELECT `id` FROM `xxxxxx`.`Player` WHERE `id` = ".$id."");

how to get a particular field from a database via php

I am trying to get four different values from my database. The session variable username and usernameto are working, but I want to get 4 different values -- two each from username and usernameto:
<?php
session_start(); // startsession
$Username=$_SESSION['UserID'];
$Usernameto= $_SESSION['UserTO'];
$db = mysql_connect("at-web2.xxxxxx", "yyyyy", "xxxxxxx");
mysql_select_db("db_xxxxxx",$db);
$result1 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
while($myrow1)
{
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
while($myrow2)
{
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
?>
Edit - just realized that you didn't tell us what wasn't working about the code you provided. Are you getting an error message or are you not getting the correct data back? You still should fix your query, but we'll need some more information to know what's wrong.
Your query statements shouldn't have "and" between the select parameters, so it should be:
Edit 2 - I just noticed that you had a while loop that you don't need, try this:
$result1 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
if (isset($myrow1)) {
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
if (isset($myrow2)) {
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
An example from the php manual echoing an html table
I don't know if you can derive what you need from this?
More specific: You can use:
$line = mysql_fetch_array($result, MYSQL_ASSOC);

How to display MySQL Select statement results in PHP

I have the following code and it should return just one value (id) from mysql table. The following code doesnt work. How can I output it without creating arrays and all this stuff, just a simple output of one value.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = map_query($query);
echo $result;
I do something like this:
<?php
$data = mysql_fetch_object($result);
echo $data->foo();
?>
You have to do some form of object creation. There's no real way around that.
You can try:
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
//$result = map_query($query);
//echo $result;
$result = mysql_query($query); // run the query and get the result object.
if (!$result) { // check for errors.
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result); // get the single row.
echo $row['id']; // display the value.
all you have is a resource, you would still have to make it construct a result array if you want the output.
Check out ADO if you want to write less.
Not sure I exactly understood, what you want, but you could just do
$result = mysql_query('SELECT id FROM table WHERE area = "foo" LIMIT 1');
list($data) = mysql_fetch_assoc($result);
if you wish to execute only one row you can do like this.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
echo $row[0];
there have been many ways as answered above and this is just my simple example. it will echo the first row that have been executed, you can also use another option like limit clause to do the same result as answered by others above.

Hot to update mysql_fetch_array

As you will see I am fetching the column, and trying to update column with new data.
$result2 line is my problem, I don't think I can add $row[0] in there. How do I do it?
$result = mysql_query("SELECT link FROM items LIMIT 3");
while($row = mysql_fetch_array($result))
{
$url=($row[0]);
$rez2 = get_final_url($url);
$result2 = mysql_query("UPDATE items SET link = $rez2 WHERE id = $row[0] LIMIT 1")
or die(mysql_error());
You should use quotes:
mysql_query("UPDATE items SET link = '{$res2}' WHERE id = $row[0]");
And it would be ideal to use mysql_escape_string() function.
So:
$rez2 = mysql_escape_string(get_final_url($url));
Also you're trying to use $row[0] as link and as id. Most likely you want $row[0] element to be an ID, and something like $row[n] where n > 0 to be a link. But if you still want to use link you should query in the following manner:
$result2 = mysql_query("UPDATE items SET link = '$res2' WHERE link = {$row[0]}");
And do not forget to escape $row
It is a good idea to use mysql_fetch_assoc() function - in this case you'll get an associative array, so you'll be able access elements by sql column names. And as result you could do something like:
$result = mysql_query("SELECT id, link FROM items LIMIT 3");
while($row = mysql_fetch_assoc($result))
{
$url=($row['link']);
$rez2 = mysql_escape_string(get_final_url($url));
$result2 = mysql_query("UPDATE items SET link = '{$res2}' WHERE id = {$row['id']}")
or die(mysql_error());
}
Also if ID is a primary key you do not need LIMIT 1 in the update query.
$row[0] is in fact valid in double-quoted strings. I think your problem is a misspelling: first you assign $rez2 a value and then in the query you use $res2.
What does get_final_url($url); do? If it doesn't surround link with quotes, and handle proper string escaping (i.e. mysql_real_escape_string), your query won't work.

Simple way to read single record from MySQL

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 :)

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