I can't seem to print the data form mysql using the following code.
Am I missing something?
<?php
mysql_connect("localhost","user","pass");
mysql_select_db("myDB");
$res=mysql_query("select * from myTABLE");
while($row=mysql_fetch_array($res))
{
echo "$row[id]";
?>
Check the PHP Manual on echo,
This part shows you how to print arrays:
// You can also use arrays
$baz = array("value" => "foo");
echo "this is {$baz['value']} !"; // this is foo !
You are missing closing curly bracket. And put id into quotation marks:
<?php
mysql_connect("localhost","user","pass");
mysql_select_db("myDB");
$res=mysql_query("select * from myTABLE");
while($row=mysql_fetch_array($res))
{
echo $row['id'];
}
?>
I recommend using mysqli. mysql extension is deprecated.
Appreciate your responses. I am able to make it work now.
I am trying to make the code more specific and secure.
I tried joining two tables, using this and it won't print the id from main (m.id) and category_name on c.category_name table.
$res = mysqli_query("SELECT m.id, c.category_name FROM `main` AS m INNER JOIN `categories` AS c ON m.category_id=c.id");
The code below works
$res = mysql_query("SELECT * from main");
Regarding connecting to the db via a separate file. This doesn't work. And no errors are being display on the php even after adding error_reporting.
require_once('../../../hide/dbconn.php');
instead of using below
mysql_connect("localhost","user","pass");
mysql_select_db("myDB");
Related
I have been using a hook in the Advanced Custom Fields plugin (load_field) which loads objects from a table in my database to an ACF select field. The table ('wp_new_royalsliders') is created by the RoyalSlider image slider plugin so i use the hook to populate a select field with the slider names.
This function has worked fine for a long time but recently stopped working - I think after updating core to 4.8.2:
add_filter('acf/load_field/name=media_gallery_slider', 'my_acf_royalslider_choices');
function my_acf_royalslider_choices($field){
$field['choices'] = array();
global $wpdb;
$query = $wpdb->prepare('SELECT * FROM %1$s ORDER BY ID ASC', 'wp_new_royalsliders');
$results = $wpdb->get_results($query);
if(!empty($results)) :
foreach($results as $result) :
$value = $result->id;
$label = $result->name;
$field['choices'][ $value ] = $label;
endforeach;
endif;
return $field;
}
When I turn debugging on I get an error:
WordPress database error: [You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%1$s ORDER BY ID ASC' at line 1]
SELECT * FROM %1$s ORDER BY ID ASC
When you pass in a hardcoded string value, you don't need placeholders, you can just use
$query = $wpdb->prepare('SELECT * FROM wp_new_royalsliders ORDER BY ID ASC');
If you want to get fancy and portable, you might opt for
$query = $wpdb->prepare('SELECT * FROM ' . $wpdb->prefix . 'new_royalsliders ORDER BY ID ASC');
So it would work on other prefixes, too, but that might not be necessary if this is a very custom thing that's not going to make it into any other site.
It appears that others have noted the removal of that possibility too, see this request.
Update: as $wpdb->prepare() expects a second parameter (since its job is to escape variable inputs for use in SQL), it might complain. Solution: get rid of it and give your SQL directly to get_results:
$results = $wpdb->get_results('SELECT * FROM ' . $wpdb->prefix . 'new_royalsliders ORDER BY ID ASC');
I've created an admin panel on my website so when the admin logs in he can edit users. I'm trying to get it to create a table that displays a list of all the users on the database, however, when I run it I get the error:
No database selected
Here is the code in my editusers.php:
<?php
include 'adminpage.php';
include 'connection.php';
$sql = "SELECT * FROM Users";
$result = mysql_query($sql)or die(mysql_error());
echo "<table>";
echo "<tr><th>UserID</th><th>First Name</th><th>Last Name</th><th>Email</th><th>D-O-B</th></tr>Username</th><th>Password</th><th>";
while($row = mysql_fetch_array($result)){
$userid = $row['UserID'];
$firstname = $row['FirstName'];
$lastname = $row['LastName'];
$email = $row['Email'];
$dob = $row['DateofBirth'];
$username = $row['Username'];
$password = $row['Password'];
// Now for each looped row
echo "<tr><td style='width: 200px;'>".$userid."</td><td style='width: 200px;'>".$firstname."</td><td>".$scale."</td><td>".$lastname."</td><td>".$email."</td></tr>".$dob."</td></tr>".$username."</td></tr>".$password."</td></tr>";
} // End our while loop
echo "</table>"
?>
First of all it looks like you are using mysql which isn't a wise move. This is because Mysql is actually deprecated and was improved to mysqli. Your problem may be to do with your database connection. You also haven't set a database. Like I said you can set an active database in your connection script. It should or could look something like this.
<?php
$conn = mysqli_connect("localhost", "root", "password", "database");
// Evaluate the connection
if (mysqli_connect_errno()) {
echo mysqli_connect_error();
exit();
}
?>
After that, your sql query is correct by selecting all from you table 'users' but in order to proceed I recommend creating a query where you use mysqli_query an select the $sql and $conn as parameters. In all honesty it is much advised to stop and continue once you have adapted to mysqli. Alternatively you can use PDO which in some cases can be seen as better to use rather than mysqli but the choice is yours. I personally would get to grips with mysqli and then look at some answers on Stack Overflow to decide whether you should use PDO or not. Visit the PHP manual here. Enter all the mysql functions you know and it will show you how to use the new mysqli version of the functions. Don't think that it is as simple as just adding and 'i' to the end of a mysql function. That's what I initially thought but there is alot to do with extra parameters etc. Hope this helps :-)
I'm using a json-parser in Xcode to fetch a table from phpmyadmin. The parser gets (or should get) the json-formated document via a php-file uploaded on my ftp-server. The file is successfully parsed but it doesn't recognize any objects. I think this is because there are multiple arrays in the json-document.
When there's only one entry the document looks like this:
[{"id":"1","Name":"Eric","Message":"first from web"}]
but when i add an entry it looks like this:
[{"id":"1","Name":"Eric","Message":"first from web"}]
[{"id":"1","Name":"Eric","Message":"first from web"},{"id":"6","Name":"Claes","Message":"Hurrburr"}]
As you can see the old array (containing only the single entry) is still there in the second array with both entries.
I suspect the problem is that the old arrays are still there when i update the database because when i tried parsing the json document with only one entry (only one array) it worked.
So my question is first if there´s something i missed in my code, or if you know why the old arrays are still there when I update the database or how to remove all previous arrays when the document is updating.
Here is my .php-file:
<?php
$username = "perhaps not sharing this information";
$password = "or this";
$database = "nah";
mysql_connect("the server url",$username, $password);
#mysql_select_db($database) or die("Error here");
$query = "SELECT * FROM debug_db";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_numrows($result);
mysql_close();
$rows = array();
while($r = mysql_fetch_assoc($result))
{
$rows[] = $r;
echo json_encode($rows);
}
?>
And check out the json-document at: http://app.levinnovation.se/getjson.php
Thank you!
i keep having this error "mysql_fetch_array() expects parameter 1 to be resource, null given in" when i try to display the returned value of count in sql. heres my code.
$query="SELECT med_rec_ID, COUNT(med_rec_ID)
FROM med_issue
WHERE MONTH(issue_date) = MONTH('2013-02-05')
GROUP BY med_rec_ID";
$result= mysql_query($query);
while($count = mysql_fetch_array($display3)){
echo $count[0];
}
i have tried to run the query in sql alone it displays 2 columns (the med_rec_ID, and the COUNT). guys how do i display the count and fix the error too?
First of all, don't use mysql_* functions since they're deprecated. Use mysqli or PDO.
Secondly, look at what you're passing into the fetch_array function.
You probably want to do something like:
$link = mysqli_connect("localhost", "admin", "pass", "db_name");
$result = mysqli_query($link, $sql);
while($row = $result->fetch_array(MYSQLI_ASSOC)){
$medIds[] = $row['med_rec_ID'];
...
}
Then fix the count by giving it an alias.
Please note that you should actually store how you access the DB in a more secure manner, but I use this only to illustrate the example. Here's a pretty good post: How to create global configuration file?
Is your query even executing? that error will happen if mysql_query doesnt return the resource, in case query fails
$query="SELECT med_rec_ID, COUNT(med_rec_ID) as C FROM med_issue where MONTH(issue_date) = MONTH('2013-02-05') GROUP BY med_rec_ID";
$result= mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
echo $row["C"];
}
Note: Please do not use mysql_* functions anymore
Give it an alias:
SELECT
med_rec_ID,
COUNT(med_rec_ID) TheCount
FROM med_issue
where MONTH(issue_date) = MONTH('2013-02-05') GROUP BY med_rec_ID
then you can select that column TheCount inside the while loop with $row['TheCount'], also use lope through the $result:
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
echo $row['TheCount'];
}
I'm trying to make a mysql query to select several tables and LEFT join them, however they all have same columns names 'user' etc. I want to rename all the fields in this manner . so I tried the following query
SELECT mod_backup_accounts . * AS account . * , mod_backup_subscriptions . *
FROM `mod_backup_accounts`
LEFT JOIN `mod_backup_subscriptions` ON `mod_backup_accounts`.subscription_id = `mod_backup_subscriptions`.package_id
However the mod_backup_accounts . * AS account . * makes it fail, is there a way to do this? so it would be names as account.
You cannot supply a shorthand to alias columns you must do it explicitly for each column name. In general anyway, it is typically recommended to name all columns explicitly in the SELECT list rather than using SELECT *, since it allows you to deterministically specify the column order, and protects you against accidentally pulling in a large BLOB later on if one ever gets added to the table ( or any other schema changes ).
SELECT
mod_backup_accounts.user AS account_user,
mod_backup_subscriptions.user AS subscription_user,
...
...
FROM
mod_backup_accounts
LEFT JOIN `mod_backup_subscriptions` ON `mod_backup_accounts`.subscription_id = `mod_backup_subscriptions`.package_id
I totally understand your problem about duplicated field names.
I needed that too until I coded my own function to solve it. If you are using PHP you can use it, or code yours in the language you are using for if you have this following facilities.
The trick here is that mysql_field_table() returns the table name and mysql_field_name() the field for each row in the result if it's got with mysql_num_fields() so you can mix them in a new array.
You can also modify the function to only add the "column." prefix when the field name is duplicated.
Regards,
function mysql_rows_with_columns($query) {
$result = mysql_query($query);
if (!$result) return false; // mysql_error() could be used outside
$fields = mysql_num_fields($result);
$rows = array();
while ($row = mysql_fetch_row($result)) {
$newRow = array();
for ($i=0; $i<$fields; $i++) {
$table = mysql_field_table($result, $i);
$name = mysql_field_name($result, $i);
$newRow[$table . "." . $name] = $row[$i];
}
$rows[] = $newRow;
}
mysql_free_result($result);
return $rows;
}