I am trying to get custom data from my wordpress database, but my knowledge in sql ist that good. Could some one help me?
Please look at my example.
I am trying to echo out image names (filename) from table (wp_ngg_pictures) that have galleryid = 2
So in this example I would get the list of :
bildes-140.jpg
bildes-127.jpg
bildes-133.jpg
bildes-152.jpg
And so on....
Assuming your using PHP all you would do is connect to your database and then do something like this
$result = mysql_query("SELECT * FROM wp_ngg_pictures WHERE galleryid = 2")or die (mysql_error());
while ($row = mysql_fetch_assoc($result)) {
echo $row['filename'];
}
If you wanted more specific results it would be a little different.
Fixed the errors.. sorry bout that, still learning myself.
This should Loop though all galleries with id of 2 and echo out the id and file name.
$GetAlbum=2;
$sql = "SELECT * FROM YOURDATABASE.wp_ngg_pictures WHERE galleryid LIKE '%".$GetAlbum."%'";
$result=mysql_query($sql) or die ('Error! Could not get gallery Databse.');
$num=mysql_num_rows($result);
$i=0;
while ($i < $num) {
$gallery=mysql_result($result,$i,"galleryid");
$galleryImg=mysql_result($result,$i,"filename");
echo 'ID '.$gallery.' File '.$galleryImg.'';
}
SELECT filename from wp_ngg_pictures WHERE galleryid = 2
Related
hi i am trying to write a php to grab a specific category of post in a wordpress database, this is what i got so far:
$q = "select * from table1 where column like 'condition'";
$r = mysql_query($q);
$id = array();
if($r){
while ($row = mysql_fetch_array($r)) {
$link = $row["object_id"];
$id[] = array(
"postid"=>$link,
);
}
}
else{
echo mysql_error();
}
now i am trying to use the result i got from pervious code to search another table to get something else. i am new to php so i am hoping to get some help
thanks in advance
I am a novice when it comes to PHP but I don't understand if my syntax is wrong in this statement, or how would I grab an int from my MySQL server.
I know that my server credentials are working fine. How would I fix this statement to give me a returned integer of the number of reviews in the userinfo table?
$numberofpreviousreviews = mysql_query("SELECT `number_of_reviews` FROM `userinfo`") or die(mysql_error()); //Check to see how many reviews user has previously created
$amountofreviews = $numberofpreviousreviews + 1;
$query2 = mysql_query("ALTER TABLE userinfo ADD `amountofreviews` VARCHAR(10000)") or die(mysql_error()); //Make another column in database for the new review
You need to fetch your results after you run your query. There are several ways to do this but using mysql_fetch_assoc() will work for you.
$numberofpreviousreviews = mysql_query("SELECT `number_of_reviews` FROM `userinfo`") or die(mysql_error()); //Check to see how many reviews user has previously created
$row = mysql_fetch_assoc($numberofpreviousreviews);
$amountofreviews = $row['number_of_reviews'] + 1;
FYI, you shouldn't be using mysql_* functions anymore. They are deprecated and going away. You should use mysqli or PDO.
Assume you have a table userinfo which has the following structure and data :
Scenario #1 :
If you want to retrieve the all number_of_reviews, then do like this,
$query = "SELECT `number_of_reviews` FROM `userinfo`";
$result = mysqli_query($db,$query);
while ($row = mysqli_fetch_assoc($result)) {
echo "Number of reviews : " . $row['number_of_reviews'] . "<br/>";
}
It will give you,
Number of reviews : 20
Number of reviews : 40
Since, the result has many rows, it will display like above.
Scenario #2:
If you want to retrieve only the specific number_of_reviews for some user id (which is unique). I take id as 1 as a example here. Then do like,
$query2 = "SELECT `number_of_reviews` FROM `userinfo` WHERE `id` = 1";
$result2 = mysqli_query($db,$query2);
while ($row2 = mysqli_fetch_assoc($result2)) {
echo $row2['number_of_reviews'] . "<br/>";
}
This will print,
20.
Because, number_of_reviews is 20 for id 1.
wish to check the last id (cat_id) in a table (category) and insert that result into a variable that I can echo.
My intention is to create a record last cat_id +1 as long as it doesnt already exist of course.
What I thought I should do was something like this;
<?php
require "mydbdetails.php";
$query="SELECT cat_id FROM category ORDER BY cat_id DESC LIMIT 1";
$result=mysql_query($query);
echo ($result);
?>
But oh no, nothing so simple. The echo was only to check I had the correct result (in phpmyadmin it returns the desired number)
Then I was hoping to be able to, with a simple html form, was to ask if the user wanted to add a category through a text box:
addrec.html:
<form action="addrec.php" method="post">
Category: <input type="text" name="category">
<input type="submit">
</form>
addrec.php:
<?php
require "mydbdetails.php";
$new_id = $result + 1;
$query="INSERT INTO category VALUES ($new_id, 'Fruits')";
?>
You must first know as a developer, that mysql extension in php will be fully deprecated in the future as it is already in the newer php versions. So use instead Mysqli extension and PDO for sanitation and securer code for your database.
As it goes to your question:- Try the following ;
// Make a MySQL Connection
$query = "SELECT cat_id FROM category ORDER BY cat_id DESC LIMIT 1";
//assign result to a variable
$result = mysql_query($query) or die(mysql_error());
//fetch result as an associative array
$row = mysql_fetch_array($result) or die(mysql_error());
echo $row['cat_id'];
You can assign it to avariable $row['cat_id'] = $catId; like that and use it .
echo $result[0]['cat_id'] i think
Please consider using PDO query Syntax instead of the old deprecated mysql_query.
If you make a PDO connection and store it in the $conn object ( pretty similar to what you already have in mydbdetails.php) just do:
$query=$conn->query("SELECT cat_id FROM category ORDER BY cat_id DESC LIMIT 1");
$result=$query->fetchAll(PDO::FETCH_COLUMN,0);
echo ($result[0]);
First of all, you should really be using mysqli instead of mysql because mysql is deprecated and will be removed in PHP 5.5.
Of course the PDO would be better, but i think that you're so new that it would be a bit to much right now.
Basically, you're firing a Query, but you don't tell the query to what connection. You're doing this:
$result=mysql_query($query);
What it should be is
$result=mysqli_query($link, $query);
Where $link is the variable where you're setting up the database connection in the mydbdetails.php file.
Without the connection the Query doesn't know where to get the data from.
But if OOP isn't new to you, the answer from #amenadiel is better because it's an OOP way.
Further, there is no need for your $new_id = $result + 1; line.
IDs should almost always set to Auto Increament in the database, so this line will be done automatically in the database when you're adding a new dataset.
More information here
Hope this helps
<?php
require "mydbdetails.php";
$query="SELECT cat_id FROM category ORDER BY cat_id DESC LIMIT 1";
$result=mysql_query($query);
$i;
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
var_dump($row); /* dump rows*/
/* or add to array to use later */
$results[$i] = $row; // add to array
$i++; // +1 the counter to increment the array
}
?>
Try to write your scripts in mysqli_ or pdo functions. Work around in mysql_ function (not recommended)
<?php
require "mydbdetails.php";
$query="SELECT cat_id FROM category ORDER BY cat_id DESC LIMIT 1";
$result=mysql_query($query);
if (mysql_num_rows($result) > 0) {
$row = mysql_fetch_array($result);
$cat_id = $row['cat_id'] + 1;
} else {
$cat_id = 1;
}
?>
cat_is is the primary key? If yes you can set that column as auto increment. You can try this
https://php.net/manual/en/function.mysql-result.php
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 $r['qty'];
-
Thanks
I can't figure out why this keeps breaking my page. Could someone take a look? Thanks!
while ($row = mysql_fetch_array(mysql_query("SELECT * FROM `mytable` WHERE `col1` = 0"))) {
echo $row['id'];
}
I've seen this type of while loop show up pretty regularly in google searches and just browsing through stackoverflow. I don't know why it isn't working for me.
How would I achieve the desired result? (Echo the id of each row where col1 = 0)
Obviously it is because you create new MySQL resource with every iteration and then use it's first row.
Use something like
$res = mysql_query("...");
while ($row = mysql_fetch_array($res)) {
echo $row['id']'
}
By the way, you do know that mysql is deprecated, do you not?
$result = mysql_query("SELECT * FROM `mytable` WHERE `col1` = 0");
while ($row = mysql_fetch_array($result)) {
echo $row['id'];
}