I have set up a MySql database containing several tables (needed for uploading tons of data that is entered in excel). I would like to use each table name as a variables that automatically updates each time a new table is added to the database. However, I don't know how to prevent PHP from overwriting my $tableName variable in loops. This is what I got so far:
<?php
if (!mysql_connect($host, $username, $password)) {
echo 'Could not connect to mysql';
exit;
}
$sql = "SHOW TABLES FROM $database";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
$tableName = $row[0];
include('../design.php');
}
?>
In other words: On the frontpage of the website I would like to include data from every single database table. I have organized a framework for extracting that data and presenting it neatly in design.php. However, this file makes use of the $tableName variable for extracting the data and I think that variable is being overwritten at each loop causing the error (only the number one table will be displayed).
Instead, if I just separate the code in their own php objects, then everything works just fine:
<?php
$tableName = tablename1;
include('../design.php');
?>
<?php
$tableName = tablename2;
include('../design.php');
?>
etc..
Can anyone explain to me how this works? Or if there is some smarter way of doing things?
Thanks in advance!
include the design.php file at the beginning of the script. Inside this file, define a function.
function doSomethingWithTableName ($table)
{
// do something with it
}
Then, in your frontend file, you would do this:
$sql = "...";
$result = mysql_query ($sql);
while ($row = mysql_fetch_row($result))
{
$processed_data = doSomethingWithTableName ($row[0]);
// do something with $processed_data
}
Hope this helps.
I've no idea what's inside design.php but instead of including it inside the loop I'd create a function, something like:
function designFunction($tableName){
//put the design.php here
return $something;
}
Now use the designFunction inside your loop
while ($row = mysql_fetch_row($result)) {
$tableName = $row[0];
$someResult = designFunction($tableName);
//etc...
}
Related
I want to echo onto my screen all the data is my 'userdata' table. i have looked around and found this code but when i run it i get a HTTP ERROR 500.
this is my code that im trying to use:
<?php
$database = new SQLite3('home.db');
$result = $database->query("SELECT * FROM userdata");
echo $result;
?>
the $database->query() method will return an SQLite3Result object which you can't just "echo". Instead, you should loop through all the results like so:
<?php
$database = new SQLite3('home.db');
$result = $database->query("SELECT * FROM userdata");
while ($row = $result->fetchArray()) {
print_r($row);
}
?>
The $row variable inside the while loop will be an array. Use the appropriate index to get the value of a single column if necessary.
Okay so I am trying to create a custom function that will echo a site url inside an iframe for the end user.
The script has to check whether or not the user has already seen the site and if they have seen it don't display it any more, but take another site url from the database etc.
Here's what I have come up with so far:
function get_urls() {
require 'config.php';
global $con;
global $currentUsername;
$con = mysqli_connect($hostname, $dbusername, $dbpassword, $dbname);
$query = "SELECT site_url FROM sites WHERE site_url IS NOT NULL";
$result = mysqli_query($con, $query);
// Get all the site urls into one array
$siteUrls = array();
$index = 0;
while($row = mysqli_fetch_assoc($result)) {
$siteUrls[$index] = $row;
$index++;
}
$query2 = "SELECT site_url FROM views WHERE user = '$currentUsername' AND site_url IS NOT NULL";
$result2 = mysqli_query($con, $query2);
// Get urls the user has already seen into another array
$seenUrls = array();
$index = 0;
while($row2 = mysqli_fetch_assoc($result2)) {
$seenUrls[$index] = $row2;
$index++;
}
// Compare the two arrays and create yet another array of urls to actually show
$urlsToShow = array_diff($siteUrls, $seenUrls);
if (!empty($urlsToShow)) {
// Echo the url to show for the iframe within browse.php and add an entry to the database that the user has seen this site
foreach ($urlsToShow as $urlToShow) {
echo $urlToShow;
$query = "INSERT INTO views VALUES ('', '$currentUsername', '$urlToShow')";
mysqli_query($con, $query);
break;
}
}
// Show the allSeen file when all the ads are seen
else {echo 'includes/allSeen.php';}
mysqli_free_result($result);
mysqli_close($con);
}
I have currently found two errors with this. First the $siteUrls and $seenUrls are all okay, but when I compare the two using array_diff then it returns an empty array.
Secondly the script doesn't write the site url into the database because the $urlToShow is an array not a single url?
I think the problem is in your code is at the place where you are creating your $siteUrls, $seenUrls arrays. mysqli_fetch_assoc() function will give you a result row as an associative array. So if you want to change some of your code in the while loops.
Please chnage this
while($row = mysqli_fetch_assoc($result)) {
$siteUrls[$index] = $row;
$index++;
}
To
while($row = mysqli_fetch_assoc($result)) {
$siteUrls[$index] = $row['site_url'];
$index++;
}
And in the second while loop also. Change this
while($row2 = mysqli_fetch_assoc($result2)) {
$seenUrls[$index] = $row2;
$index++;
}
To
while($row2 = mysqli_fetch_assoc($result2)) {
$seenUrls[$index] = $row2['site_url'];
$index++;
}
and try
i would not run two queries and try to merge the result array. you can use mysql itself to just return the sites that have not been seen yet:
SELECT sites.site_url FROM sites
LEFT JOIN views ON views.site_url=sites.site_url AND views.user='$currentUsername'
WHERE sites.site_url IS NOT NULL AND views.site_url IS NULL
This will return only site_urls from sites, that have no entry in the views table. The LEFT JOIN will join the two tables and for every non-matching row there will be NULL values in the views.site_url, so that is why i am checking for IS NULL.
Your saving of $urlToShow should work, if you set to $row field content and not $row itself as suggested, but if you want to check what is in the variable, don't use echo use this:
print_r($urlToShow);
If the variable is an array, you will see it's content then.
#Azeez Kallayi - you don't need to index array manually.
$seenUrls[] = $row2['site_url'];
In addition, you can fetch all result
$rows = mysqli_fetch_all($result2,MYSQLI_ASSOC);
foreach($rows as $row){
echo $row['site_url'];
}
I've got several queries I want to run on a single page. I obviously don't want to put the actual queries in my template file, so I think what I want to do is construct a function and call it wherever I want the query results to show up. Right?
So, for example, I'll have <?php sidebar_query()?> in the sidebar, <?php footer_query()?> in the footer, etc.
So, I just make a file called functions.php, do PHP include, and put something like this in there?
<?php
function sidebar_query(){
$query = ("SELECT sidebarposts FROM table;");
return $query;
}
?>
or do you use echo and not return anything?
<?php
function sidebar_query(){
$query = ("SELECT sidebarposts FROM table;");
echo $query;
}
?>
Along the exact same line, I'd like to count the results that get returned, and display a 'There were X posts returned!' message below. Seems like it would make sense to put this in a function too. How would I make a 'generic' function that I could reuse for each query?
<?php
function number_of_results(){
$num_rows = mysql_num_rows();
echo $num_rows;
}
?>
I'd be extremely grateful if someone could give me the theoretical gist of what I should be trying to achieve here.
Thanks for helping a beginner.
Terry
I think I get what you mean.
Return the value instead like this
function sidebar_query(){
$rValue = "";
$query = ("SELECT sidebarposts FROM table;");
$result = mysql_query($query);
if ($row = mysql_fetch_array($result)){
$rValue = $row['sidebarposts'];
}
return $rValue;
}
Now you can echo sidebar_query(); or whatever you want to do with it.
By doing ("SELECT sidebarposts FROM table;")you're not actually doing anything, you just have a string stored as a variable.
A simple example is
function sidebar_query()
{
$query = mysql_query("SELECT title,post FROM table;"); //query the db
$resArr = array(); //create the result array
while($row = mysql_fetch_assoc($query)) { //loop the rows returned from db
$resArr[] = $row; //add row to array
}
return $resArr;
}
Then to use it you can do
$sideBarPosts = sidebar_query(); //get the result array
foreach($sideBarPosts as $post) { //loop the array
echo '<h1>'. $post['title']. '</h1>';
echo '<p>'. $post['post']. '</p>';
}
EDIT. I see you want to let the function print it directly, you can do that instead of returning the array, if you like.
function sidebar_query(){
$query = mysql_query("SELECT * FROM table;");
$result = $conn->query($query);
$resArr = array(); //create the result array
while($row = $result->fetch_assoc()) { //loop the rows returned from db
$resArr[] = $row; //add row to array
}
return $resArr;
}
and that :
$sideBarPosts = sidebar_query(); //get the result array
foreach($sideBarPosts as $post) { //loop the array
echo '<h1>'. $post['title']. '</h1>';
echo '<p>'. $post['post']. '</p>';
}
// Set Connection with database
$conn = mysql_pconnect($servername,$username,$password) or die("Can not Connect MySql Server!");
// Select database you want to use
mysql_select_db($databasename,$conn) or die("Can not select Data Base!");
// When you want to query SQL inside php function, you need to pass connection($conn) to the function as below.
function sidebar_query($condb)
{
$strSql = "SELECT sidebarposts FROM table";
$result = mysql_query($strSql,$condb);
return $result;
}
// Call function
sidebar_query($conn);
This is work for me as i use with my webpage.
Read the php doc on how to run mysql query.
Yay you have the query, now you need to do something with it:
mysql_query($query); for example might work for you :-)
You want to run the query and return the result in whatever format you want to use in your template.. so for the raw number you want to return an integer, for the sidebar posts return an array of posts OR the mysql result.
You should make the "'There were X posts returned!' " thing a completely different function which you can pass in an integer an an optional string. This way you can reuse it for a ton of stuff. for example:
function format_nb_records($nb_records, $format = 'There were %s posts returned!'){
return sprintf($format, $nb_records);
}
Although in this case i dont think there is really enough extra logic to warrant wrapping it in a function. I would probably just do this directly in my template.
A good example of where something like this is useful is if you want to do "time in words" functionality like "Yesterday" or "10 minutes ago" then you would pass in the time calculate which string to use, format it and return it.
Although in this case i dont think there is really enough extra logic to warrant wrapping it in a function. I would probably just do this directly in my template.
A good example of where something like this is useful is if you want to do "time in words" functionality like "Yesterday" or "10 minutes ago" then you would pass in the time calculate which string to use, format it
You need to pass database connection name to function like this,
You already have a connection name, mine is $connect_name
$connect_name= mysql_connect( 'host', 'user', 'password');
function getname($user,$db){
$data= mysql_query("SELECT $user...", $db);
//your codes
//return bla bla
}
echo getname("matasoy",$connect_name);
This work for me.
I also use this for medoo.min sql system. You can use medoo with functions like this;
function randevu($tarih,$saat,$db){
$saat_varmi = $db->count("randevu",array("AND"=>array("tarih"=>$tarih,"saat"=>$saat)));
if($saat_varmi>0)
return 3;
else
return 2;
}
have a nice day, Murat ATASOY
I'm writing a PHP file which will putt from multiple table rows, data which it must echo on one page.
I wrote a function command because I a loop wont work as the information isn't outputting in the same manner every time.
here is how it looks so far.
include("dbconnect.php");
function get_table_data($id)
{
$row = mysql_query('SELECT row FROM table WHERE id = '.$id.'');
$row2 = mysql_fetch_array($row);
};
$sss = 18; // this is the ID
echo get_table_data($sss);
My issue is that nothing echos and I have no errors
first of all, you need to return something:
function get_table_data($id) {
$row = mysql_query('SELECT row FROM table WHERE id = '.$id.'');
return mysql_fetch_array($row);
}
then, you need to access the row:
$result = get_table_data($sss);
echo $result['row'];
Hope this helps!
Your function get_table_data should return some value to be echoed in your statement echo get_table_data($sss);
Ok i got a problem now i want to display a data from the database and display it through a function now how do i do that??
like i have fetched a row from the database and its name is $row_field['data']; and it is correct now i have assigned a variable to it like this $data = $row_field['data']; now if i call it in a function it shows undefined variable even after i assigned it global in the function like this
function fun(){
global $data;
echo $data;
}
but if i assign it a value like 1 or 2 or anything it gets displayed without any error why is that so??
If it displays if you assign it a value like 1 or 2 while still in the global scope, then I can only assume that your database did not return the result you thought it did. Does the database value display if you echo it out outside of the function?
Global is evil. I dont know what you are trying to do, but why dont you just do the query in the function itself?
If you have a column named data and your php call was something like
$result = mysql_query("SELECT data FROM mytable");
while ($row_field = mysql_fetch_assoc($result, MYSQL_NUM)) {
...
}
Then you could replace ... with print $row_field['data'].
Else please provide a snippet of your code where you query the db and retrieve the result.
When learning php try to start with simple things. For example in order to get some data from a database follow the examples from php website.
<?php
$conn = mysql_connect("localhost", "mysql_user", "mysql_password");
if (!$conn) {
echo "Unable to connect to DB: " . mysql_error();
exit;
}
if (!mysql_select_db("mydbname")) {
echo "Unable to select mydbname: " . mysql_error();
exit;
}
$sql = "SELECT id as userid, fullname, userstatus
FROM sometable
WHERE userstatus = 1";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
// Note: If you put extract($row); inside the following loop, you'll
// then create $userid, $fullname, and $userstatus
while ($row = mysql_fetch_assoc($result)) {
echo $row["userid"];
echo $row["fullname"];
echo $row["userstatus"];
}
mysql_free_result($result);
If all this goes well go a little further change a little the while loop.
$myArray = array();
while ($row = mysql_fetch_assoc($result)) {
$myArray[] = $row;
}
mysql_free_result($result);
// now you can start playing with your data
echo $myArray[0];
Small steps...