I have simple code for displaying images. I created table with 4 columns (ID, location, capture, equence) and inserted there 18 records. My question is: how to display all records from table in reverse mode? I need to make that the last entry will be displayed first, and the first entry displayed last.
What I need: 18-1
What I have now: 1-18
I was searching for simple codes to do that, but notwing worked at all. So i'd be very grateful if someone will help me to solve that problem.
Heres the basic code of my display script:
<?php
mysql_connect("localhost", "***", "***") or die(mysql_error());
mysql_select_db("martinidb1337") or die(mysql_error());
$result = mysql_query("SELECT * FROM klpgalerija") or die(mysql_error()); while($row = mysql_fetch_array( $result )) {
echo '<p><img src="'.$row['location'].'"></p>';
}
You have to use MySQL ORDER BY clause for that,
SELECT * FROM klpgalerija ORDER BY id DESC
Note: Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated.
So use either PDO or MySQLi (IMO PDO is way to go)
Changed query from "SELECT * FROM klpgalerija" to "SELECT * FROM klpgalerija ORDER BY ID DESC"
<?php
mysql_connect("localhost", "***", "***") or die(mysql_error());
mysql_select_db("martinidb1337") or die(mysql_error());
$result = mysql_query("SELECT * FROM klpgalerija ORDER BY ID DESC") or die(mysql_error()); while($row = mysql_fetch_array( $result )) {
echo '<p><img src="'.$row['location'].'"></p>';
}
add an order by desc clause in your sql query
$result = mysql_query("SELECT klpgalerija.* FROM klpgalerija order by klpgalerija.ID desc") or die(mysql_error());
Related
I'm trying to get ORDER BY id DESC to work in this line of php but I can't get it to work. It works without it though but they just display in the opposite order. Where should I position it?
$query = mysql_query("SELECT * FROM photos ORDER BY id DESC WHERE title LIKE '%".$search."%'");
Update:
I've updated the php with your suggestion that works now thanks. I've also updated it to mysqli as suggested, could this be structured better in anyway? It is working, just wondered if anyone has any improvements?
<?php
$search = $_GET['search'];
$db = mysqli_connect("", "", "", "") or die ("could not connect to mysql");
$sql = "SELECT * FROM photos WHERE title LIKE '%".$search."%' ORDER BY id DESC";
$result = mysqli_query($db, $sql);
if(mysqli_num_rows($result) >=1) {
while ($row = mysqli_fetch_array($result)) {
echo"<div id='img_div'>";
echo"<img src='images/".$row['image']."'>";
echo"<h1>".$row['title']."</h1>";
echo"<p>".$date = date('j F, Y', strtotime($row['date']))."</p>";
echo"<p>".$row['link']."</p>";
echo"</div>";
}
//continue
}else{
echo "No Results";
}
?>
Your query is wrong. ORDER BY id DESC should be placed after the WHERE clause, like this:
$query = mysql_query("SELECT * FROM photos WHERE title LIKE '%".$search."%' ORDER BY id DESC");
Sidenote(s):
Don't use mysql_* functions, they are deprecated as of PHP 5.5 and are removed altogether in PHP 7.0. Use mysqli or pdo instead. And this is why you shouldn't use mysql_* functions.
Learn about prepared statement because right now your query is susceptible to SQL injection attacks. Here's a good read on how you can prevent SQL injection in PHP.
I am trying to display all results from my database where the username is equal to the person that is logged in and where the search row isn't empty.
Here is the code that I thought up but it doesn't work. Does anyone have any ideas on how I could do this?
<?php $result = mysql_query("SELECT * FROM searchresults WHERE user='{$_SESSION['user_name']}' WHERE search!='' ORDER BY date DESC") or die(mysql_error()); ?>
The issue I am having is that I don't know how to do more than one WHERE in the query.
Use AND here
<?php $result = mysql_query("SELECT * FROM searchresults
WHERE user='{$_SESSION['user_name']}' AND search!='' ORDER BY date DESC") or die(mysql_error()); ?>
Use AND keyword instead of the where. There is only one where clause is allowed in single query.
<?php $result = mysql_query("SELECT * FROM searchresults WHERE user='{$_SESSION['user_name']}' AND search!='' ORDER BY date DESC") or die(mysql_error()); ?>
use the below code
<?php
$result = mysql_query("SELECT * FROM searchresults WHERE user='{$_SESSION['user_name']}' and search!='' ORDER BY date DESC") or die(mysql_error());
?>
I want to show the maximum value of a particular column of mysql table in php. Here is my code:
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("database") or die(mysql_error());
$var = $_POST['value'];
$sql = mysql_query(" SELECT MAX(column) FROM tableName WHERE variable='$var' ") or die(mysql_error());
$row = mysql_fetch_array($sql) or die(mysql_error());
echo "Maximum value is :: ".$row['column'];
?>
output:::
Maximum value is ::
Or you could be a bit creative:
$sql = mysql_query("SELECT `column` FROM tableName WHERE variable='$var' ORDER BY `column` DESC LIMIT 1") or die(mysql_error());
This is the same problem that I was searching for a solution. The approach below worked. The solution was using "MYSQLI_NUM". While the code below gives the intended result it still seems way too complex. Is there a way to "short-cut" using the "while" statement to search an array since there is only one returned value?
Selecting a max value only returns one number, yet PHP still seems to require a loop to evaluate an entire array. I was expecting something easy, like: max_value=result(0).
<?php
require_once 'login.php';
$conn = new mysqli($hn, $un, $pw, $db);
if ($conn->connect_error) die($conn->connect_error.' Sorry, could not connect to the database server');
$query = "SELECT MAX(IssueIDNUM) FROM tblIssueList";
$result =$conn->query($query);
if (!$result) die($conn->error);
while($row=mysqli_fetch_array($result, MYSQLI_NUM))
{
$maxresult=$row[0];
echo $maxresult;
}
$result->close();
$conn->close();
?>
With further experimentation, I was also able to adapt the code by Jim H. to develop the following solution. Though it works the code still seems too complex.
// Solution #2
$result =$conn->query("SELECT MAX(IssueIDNUM) `max` FROM tblIssueList");
if (!$result) die($conn->error);
while($row=mysqli_fetch_array($result, MYSQLI_ASSOC))
{
echo $row['max'];
}
The real name of column is MAX(column), try:
print_r($row);
You may either do:
$sql = mysql_query(" SELECT MAX(column) AS `column` FROM tableName WHERE variable='$var' ") or die(mysql_error());
Or:
$row = mysql_fetch_row($sql) or die(mysql_error());
echo "Maximum value is :: ".$row[0];
You have two choices. By Index or by Column Name. If you really want to use a column name, alias the column in your query.
$sql = mysql_query(" SELECT MAX(column) `max` FROM tableName WHERE variable='$var' ") or die(mysql_error());
Then you can use
$row['max'];
or
$row[0];
You are probably not returning any rows. You should try WHERE variable LIKE '%$var%';
$sql = mysql_query(" SELECT MAX(column) AS GIVE_A_NAME FROM tableName WHERE variable='$var' ") or die(mysql_error());
and
echo "Maximum value is :: ".$row['GIVE_A_NAME'];
I want to display some basic data from a MySQL database. Here's the current code I have, but it doesn't seem to work. Could someone please explain why this doesn't work and offer a solution? Thanks!
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("cede") or die("Couldn't find database");
$result = 'SELECT * FROM 'users' ORDER BY 'DATE' DESC LIMIT 8';
echo = "'$result'"
?>
Providing your connection and structure information is correct, the following should work for you:
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("cede") or die("Couldn't find database");
$result = 'SELECT * FROM `users` ORDER BY `DATE` DESC LIMIT 8';
$query = mysql_query($result) or die("Query Error");
while($row = mysql_fetch_assoc($query))
{
echo = "'" . $row['user'] . "'";
}
?>
You forgot to query the database!
You need to use mysql_query() to retrieve data from your DB server, then loop through it with a while() loop.
Also, you can't use quotes inside quoted strings - it breaks the string, meaning you'll get a syntax error with the SELECT ... line. You don't actually need to quote database fields in queries, so the following should work fine:
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("cede") or die("Couldn't find database");
$query = 'SELECT * FROM users ORDER BY DATE DESC LIMIT 8';
$result = mysql_query($query); // Query the database.
// Loop through each returned row
while($row = mysql_fetch_assoc($result))
{
print_r($row); // Prints the current row
}
?>
To show any errors that PHP reports, put these two lines at the top of your script.
error_reporting(E_ALL);
ini_set('display_errors', '1');
They will output any errors you get, making problems much easier to solve.
After the selecting the database need
$stmt = mysql_query("SELECT * FROM users ORDER BY DATE DESC LIMIT 8");
while ($result = mysql_fetch_array($stmt, MYSQL_NUM))
{
var_dump($result);
}
mysql_free_result($stmt);
You should query your string and then echo the result, like this for example:
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("cede") or die("Couldn't find database");
$query = 'SELECT * FROM `users` ORDER BY `DATE` DESC LIMIT 8';
$result = mysql_query($query);
echo = "'$result'"; // This may need a foreach loop
?>
You should escape the ' in your string, because you use them to open and close your string too. The syntax highlighter actually tells you that you are wrong ('users' and 'DATE' are black instead of maroon). :)
Please see the PHP.net documentation about strings:
After that, you'll need to further process $result. It is just a resource pointer and cannot be echoed that way. But that's a second step. :)
<?php
mysql_connect("localhost", "user", "password") or die(mysql_error());
mysql_select_db("jmvarela_jacket") or die(mysql_error());
$query = 'SELECT * FROM `quote` ORDER BY `id` DESC LIMIT 1';
$row = mysql_fetch_array( $query );
echo $row['frase'];
?>
I cant get this to work.
I get this error:
Warning: mysql_fetch_array(): supplied
argument is not a valid MySQL result
resource in
/home/jmvarela/public_html/ihateyourjacket.com/latest/index.php
on line 7
I am trying to select the latest entry to the mysql database.
The table is called "quote"
There are three fields: id, frase and name.
Just to clarify (because this could be VERY bad coding) I am trying to get the "biggest" id and to display it's correspondant "frase".
you have not perform your query
$result = mysql_query($query);
$row = mysql_fetch_array( $result );
try this
Looks like you are not running the query.
// construct the query.
$query = 'SELECT * FROM `quote` ORDER BY `id` DESC LIMIT 1';
// run the query..THIS IS MISSING.
$result = mysql_query($query);
Also it's better to change SELECT * to SELECT frase, since you're interested only in the frase column. This will not bring all the unwanted columns from MySql to PHP, making your program perform better.
I´m not sure if this should be done but ill leave the complete running code for future refence.
<?php
mysql_connect("localhost", "user", "password") or die(mysql_error());
mysql_select_db("jmvarela_jacket") or die(mysql_error());
// construct the query.
$query = 'SELECT * FROM `quote` ORDER BY `id` DESC LIMIT 1';
$result = mysql_query($query);
$row = mysql_fetch_array( $result );
echo $row['frase'];
?>
Thanks to everyone!