How to get the first ID with SQL - php

I am trying to get the first id with a specific date.
So this is my code:
$checkm = "SELECT FIRST(id) FROM times WHERE date='2014-03-07'";
$resultm = mysqli_query($con,$checkm);
I have a table called times and there are some columns. Two of them are date and id.
I am trying to get the first row's id with the date 2014-03-07.
EDIT: Fixed!
$checkm = "SELECT id FROM times WHERE date='2014-03-06' ORDER BY id ASC LIMIT 1";
$resultm = mysqli_query($con,$checkm);
while($row = mysqli_fetch_array($resultm)) {
$resultm1 = $row['id'];
}

You probably want the minimum id.
SELECT min(id)
FROM times
WHERE date = '2014-03-07'

pretty straightforward...
SELECT id FROM times WHERE date='2014-03-07' ORDER BY id ASC LIMIT 1

Related

How to get the minimum id from the table in MySql

I have written a MySql query to get the columns related with minimum id . Looks something like this
SELECT min(id) as ID,feed , idpropiedad FROM `registrofeed` WHERE feed=21
The table has 4 rows looks like this
So according to the function that I have written
function setLC()
{
$sql = "
SELECT min(id) as ID
, feed
, idpropiedad
FROM `registrofeed`
WHERE feed=21
";
$result = $this->localDb->execute($sql);
$row=mysql_fetch_array($result);
echo $sql;
echo $row['idpropiedad'];
$this->lastCode = $row['idpropiedad'];
}
It returns empty string for idpropiedad
Can any one help me out where I am going wrong
Thanks in advance
I'd think the query you're actually looking for is this:
SELECT id, feed, idpropiedad
FROM registrofeed
WHERE feed = 21
ORDER BY id ASC
LIMIT 1
MIN() is giving you the generally lowest value in the column, it does not affect the rest of the columns. If you want the whole row with the lowest id it doesn't help.
To illustrate, if you really wanted to use MIN here, you'd have to do:
SELECT id, feed, idpropiedad
FROM registrofeed
WHERE id = (SELECT MIN(id) FROM registrofeed WHERE feed = 21)
You can do a better query like this:
$sql = "
SELECT id as ID
, feed
, idpropiedad
FROM `registrofeed`
WHERE feed=21
HAVING MIN(id)
";
This will return only one row with the minimum id number. It's more readable than using ORDERING AND LIMIT 1.
try your select query as
SELECT * FROM registrofeed WHERE feed='21' ORDER BY id ASC LIMIT 1
this fetches the row having minimum id.
Hope it helps
Try this
$sql = "SELECT min(id) as ID,feed , idpropiedad FROM `registrofeed` WHERE feed='21' order by id asc";

PHP, SQL - getting fetch where table id = user id and count other table where row is = user id

Thanks for helping, first I will show code:
$dotaz = "Select * from customers JOIN contracts where customers.user_id ='".$_SESSION['user_id']."' and contracts.customer_contract = ".$_SESSION['user_id']." order by COUNT(contracts.customer_contract) DESC limit $limit, $pocetZaznamu ";
I need to get the lists of users (customers table) ordered by count of contracts(contracts table)
I tried to solve this by searching over there, but I can't... if you help me please and explain how it works, thank you! :) $pocetZanamu is Number of records.
I need get users (name, surname etc...) from table customers, ordered by number of contracts in contracts table, where is contract_id, customer_contract (user id)..
This should do it where is the column name you are counting.
$id = $_SESSION['user_id'] ;
$dotaz = "Select COUNT(`customer_contract`) AS CNT, `customer_contract` FROM `contracts` WHERE `user_id`=$id GROUP BY `customer_contract` ORDER BY `CNT` DESC";
Depending on what you are doing you may want to store the results in an array, then process each element in the array separately.
while ($row = mysqli_fetch_array($results, MYSQL_NUM)){
$contracts[$row[1]] = $row[0];
}
foreach ($contracts AS $customer_contract => $count){
Process each user id code here
}
Not sure what you are counting. The above counts the customer_contract for a table with multiple records containing the same value in the customer_contract column.
If you just want the total number of records with the same user_id then you'd use:
$dotaz = "Select 1 FROM `contracts` WHERE `user_id`=$id";
$results = $mysqli->query($dotaz);
$count = mysql_num_rows($results);

How to retrieve previous record from database

I am writing a Custom Query in WordPress database to get the previous record from the posts table.
Example:
I have an ID of 34975; after I query the database I should get the ID as 34972, which is the previous record ID.
SQL
$results = $wpdb->get_results( "SELECT * FROM agencies_posts WHERE ID = '34975 ' LIMIT 1", OBJECT );
foreach( $results as $item ){
$previous_depature_port = $item->ID;
}
If I'm understanding your question correctly, you need to add ORDER BY and use < instead of =:
SELECT *
FROM agencies_posts
WHERE ID < 34975
ORDER BY ID DESC
LIMIT 1
Pretty sure you want:
select *
from agencies_posts
where id = (select max(id) from agencies_posts where id < '34975')
If the 'current' id is what's known and you just want the one prior.
Select everything with an id less than the one you are interested in, and only grab the first one
SELECT *
FROM agencies_posts
WHERE ID < '34975 '
ORDER BY ID DESC LIMIT 1"

SELECT * FROM table_name ORDER BY column_name?

ok so i coded in a news section and everytime i insert new news,its shows below the old one.
i want to make it ORDER BY id but make it start like backwards.i dont know how to explain but yeh
i want them to be ordered by the newest added by the id so if the 1st row that was inserted's id is 1 then i want it to show below the next id.
so row with id = 2 will be here
so row with id = 1 will be here
thats how i want it to be, instead of it being like this
so row with id = 1 will be here
so row with id = 2 will be here
.
Sorry for my bad explanation i hope this is understandable
heres my code so far
<?php
require("include/config.php");
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id");
while($row = mysql_fetch_array($sqlnews)) {
$dbdate = $row['date'];
$dbnews = $row['news'];
echo "<h1><strong>$dbdate</strong></h1>";
echo "<div class='content'>$dbnews</div><br><br>";
}
?>
add DESC in your ORDER BY clause
SELECT * FROM news ORDER BY id DESC
by default, it is in ASC mode.
SELECT * FROM news ORDER BY id DESC
DESC is the descending keyword ASC is ascending
If you specify neither then default behaviour is ascending
Just use DESC keyword in your sql query.
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id DESC");
However, it isn't really such a good idea to use id, because semantically, there is nothing preventing somebody from changing the sequence which counts up automatically assigning the ID.
Therefore, you should add a column created_at. Everytime you insert a row, you can use the SQL function NOW().
The advantage is that you can say:
SELECT * FROM news WHERE created_at <= NOW() ORDER BY created_at DESC
This means that you can schedule news items ahead of time, and it will automatically display when the date/time arrives!
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id DESC");
try this:
just have to add
order by id DESC
Just replace your code by this code:
<?php
require("include/config.php");
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id DESC");
while($row = mysql_fetch_array($sqlnews)) {
$dbdate = $row['date'];
$dbnews = $row['news'];
echo "<h1><strong>$dbdate</strong></h1>";
echo "<div class='content'>$dbnews</div><br><br>";
}
?>

Find out currentid next row from current id

I'm trying to select next row from current id.
How can i find out my current id?
$qry = mysql_query("SELECT performerid,pic0 FROM ".$table." WHERE id > $currentid ORDER BY id ASC LIMIT 1");
Select it?
$qry = mysql_query("SELECT id,performerid,pic0 FROM ".$table." WHERE id > $currentid ORDER BY id ASC LIMIT 1");
MySQL doesn't work with any concept of current ID. If you're looking for the last inserted ID then it can be done via mysql_inserted_id function.

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