I have a MySQLi query that returns all of the "assets" assigned to an employee based on their EmployeeID. This works great. The problem I'm facing is in the presentation.
I have an HTML table that has two sections: 1 for Hardware and 1 for software. What I am hoping to avoid is having to perform separate lookups that generate separate result sets for each type of asset. The end result needs to display as follows:
I can build the table just fine. The result sets contains a field of asset_type but I've not had any luck figuring out the code to use to iterate through my single result set. Is this even possible? Can I pull just the hardware assets from the result set with a while? Perhaps a
while($result['asset_type'] == "hardware"){
echo ""; // table row code
}
And then repeat the same thing later in my table code for asset_type software?
UPDATE 1
The code I've thought might work so far, but isn't doing anything, is
// SQL query
$q = "SELECT * FROM `assets_table` WHERE `emp_id` = '".$emp_id."'";
$r = mysqli_query($connect, $q);
$total_assets = mysqli_num_rows($r);
while($r){
if($r['category'] = "hardware"){
echo $r['asset_name']." - ".$r['hw_make']." ".$r['hw_model'];
}
}
I ended up going ahead and breaking out the query into multiple result sets and dealing with them that way. It'd be awesome if you could have a while() with a WHERE statement when iterating through result sets / arrays.
Related
Function render makes website 500% slow! Can anyone fix that please ?
Someone told me :
because it sends a database request on each iteration of the loop (it's not the only problem with this chunk of code but it's the most taxing one)
Yes I understand what that means. His way is:
you need to get all of the data before you start building the menu,
then you just insert the data instead of requesting more data on each
iteration
But i don't know how i must do it!
<?php
$menu_html='';
function render_menu($parent_id,$actmenuid)
{
$obj = new Database();
$con = $obj->dbconnectt();
global $menu_html;
$result=mysqli_query($con, "select * from tbl_menu where parent_id='$parent_id'");
if(mysqli_num_rows($result)==0) return;
if($parent_id==0){
$menu_html.='<ul class="topnav">';
}else{
$menu_html.='<ul>';
}
while($row=mysqli_fetch_array($result)) {
$childnum = $obj->recordcount("SELECT * FROM tbl_menu WHERE parent_id='".$row['id']."'");
if($childnum == 0){
$linkvalue='/category/'.$row['id'].'.html';
} else{
$linkvalue='#';
}
if($row['id']==$actmenuid && $actmenuid !=NULL){
$actv='class="active"';
}else{
$actv='';
}
$menu_html.='<li '.$actv.'>'.$row['title'].'';
render_menu($row['id'],$actmenuid);
$menu_html.='</li>';
}
$menu_html.='</ul>';return $menu_html;
}
if($isDsh==false){
echo render_menu(0,$actmenuid);
}
?>
Depending on how many records you have, try removing this query from inside the loop since it's running for every record on the first query.
$childnum = $obj->recordcount("SELECT * FROM tbl_menu WHERE parent_id='".$row['id']."'");
Change it a single query like this where it returns counts for each parent idea, and place it outside of the loop:
$parentcount = mysqli_query($con, ("SELECT parent_id, count(*) FROM tbl_menu GROUP BY parent_id");
There may be other issues, so please post the database structure and number of records that you're working with too.
Don't make recursive queries.
Having "more than 1000" rows is not too big. You can simply call everything from the table into php, then perform the recursive html build in php this will have a memory overhead, but far less processing overhead because you only ever make one trip to the db.
Alternatively (when your db table is prohibitively large), you should avoid gathering rows unnecessarily by adding a new column. The new column will store all "descendants" for the respective row when the row is INSERTed or update it when it is UPDATEd. Then you only need to reference this column when needing to call specific rows. In other words, do the recursive processing only once (when writing to the db) AND not when needing to display the data. This will, again, produce a finite result set in one query which can then be recursively traversed to build the desired output.
basically you need to do what #spudly has suggested.
But there is a small catch in his solution which depending on the number of the rows in yous tbl_menu table you may use a big chunk of memory to fetch all the records.
you can optimise it more with using his solution but changing the query to:
select
parent_tbl_menu.id,
count(child_tbl_menu.id) as cnt
from
tbl_menu as parent_tbl_menu
left join
tbl_menu as child_tbl_menu
on parent_tbl_menu.id = child_tbl_menu.parent_id
where
parent_tbl_menu.parent_id = ?
group by
parent_tbl_menu.id
This way you will only fetch the child records of a specific parent.
And please consider using prepared statements as your code has sql injection vulnerability.
Connect (from PHP to MySQL) only once for the entire web page.
Don't put a SELECT inside a loop if you can do all the work in a single SELECT, such as with a JOIN. (Exception: A "hierarchical" table needs the nested SELECT. Exception to the exception: MySQL 8.0 and MariaDB 10.2 can do it with a "recursive CTE".)
Don't fetch all the columns (SELECT *) when all you want it is a recordcount. Instead, SELECT COUNT(*) ... and use the number returned.
1000 of anything is probably excessive for a web page. Re-think the UI.
I have two mysql tables. I want to get the data from these tables and show it in a loop. The data is totally unrelated to each other and should stay that way. I just need to show data from these two different tables in the same place.
I tried the mysqli_multi_query, but I couldn't show the results from an individual column like I can with a normal query.
For each of these two tables, I need 2 SELECT statements with two WHERE clauses. Does anyone know how to do this?
I've tried all different ways of trying to get the info from both tables and just show them in one loop. I've tried mysqli_multi_query, but don't know how to save specific column results in a variable.
$sql = "SELECT *
FROM misc
WHERE height LIKE '$height_input'
SELECT *
FROM bolts
WHERE name LIKE '$bolt_name_input'
;
";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$height_row = $row['height'];
$bolt_name_input = $row['name'];
?>
Height Row: <?php echo $height_row; ?>m<br />
bolt Name: <?php echo $bolt_name_input; ?><br />
} //while
My error message is generally "Trying to get property 'num_rows' of non-object".
Yes everyone, I have taught myself PHP and MySQL through online tutorials and have not had a lot of experience with them yet. I apologise, I am still getting the hang of this.
I have just worked out now by trial and error that I can show information from a table outside of the while loop. I had no idea I could do this! Which changes everything. So I can select information from various database tables using different SELECT statements and then echo them all into one place in my HTML table underneath it on my web page. The results don't have to be shown in the while loop, they can be outside it.
So my structure now looks like this basically:
SELECT from table 1 column 1 WHERE (form input value)
num rows
while loop
$height = $row_height['height'];
end num rows
end while loop
SELECT from table 2 column 1 WHERE (form input value)
num rows
while loop
$bolts = $row_bolts['bolts'];
end num rows
end while loop
<table>
<tr>
<td>echo $height</td>
<td>echo $bolts</td>
</tr>
</table>
So, this is working for me to retrieve multiple results from my different tables. I have to change the variable names of the sql statement each time, I guess I could do it with a function so i can repeat it and make it look neater, but this is working for me this way. So, with bolts I am using the variable names $sql_bolts, $result_bolts and $row_bolts and with height I am using variable names $sql_height, $result_height and $row_height and so on.
$bolts_input = $_POST['bolts'];
$sql_bolts = "
SELECT *
FROM bolts
WHERE name LIKE '$bolts_input'
;";
$result_bolts = $conn->query($sql_bolts);
if ($result_bolts->num_rows > 0) {
// output data of each row
while($row_bolts = $result_bolts->fetch_assoc()) {
$bolts = $row_bolts['name'];
} //bolts while
} // bolts num rows
I am not using mysql, but mysqli. I did a whole bunch of tutorials on PDO and how to connect to the database and retrieve information and just couldn't figure out how to show values from a row like I can with the above. It's frustrating because I want to use the latest methods, but I can't find how to online that makes sense to me.
Thank you for eveyone's comments so far.
I am trying to figure out how to delete all ids in the database that do not exist in an array. I have been trying to use NOT IN in my query but I am not sure why it wont work when running it in a script the same way it works when I manually enter it into mysql. Here is an example.
mysqli_query($con, "DELETE FROM table WHERE id NOT IN ($array)");
$array is a list of ids from a json api. I use CURL to fetch the ids and I am trying to delete all ids in the database that do not match the ids in $array.
First I use another simple CURL script to scrape the apis and insert the ids found into the database and what I am trying to do here is basically make a link/data checker.
If the ids in the database are not found in the array when rechecking them then I want them deleted.
I thought that the query above would work perfect but for some reason it doesn't. When the query is ran from a script the mysql log shows the queries being ran as this.
Example:
DELETE FROM table WHERE id NOT IN ('166')
or this when I am testing multiple values.
DELETE FROM table WHERE id NOT IN ('166', '253', '3324')
And what happens is it deletes every row in the table every time. I don't really understand because if I copy/paste the same query from the log and run it manually myself it works perfect.
I have been trying various ways of capturing the array data such as array_column, array_map, array_search and various functions I have found but the end result is always the same.
For right now, just for testing I am using these 2 bits of code for testing 2 different apis which gives me the same sql query log output as above. The functions used are just a couple random ones that I found.
//$result is the result from CURL using json_decode
function implode_r($g, $p) {
return is_array($p) ?
implode($g, array_map(__FUNCTION__, array_fill(0, count($p), $g), $p)) :
$p;
}
foreach ($result['data'] as $info){
$ids = implode_r(',', $info['id']);
mysqli_query($con, "DELETE FROM table WHERE id NOT IN ($ids)");
}
And
$arrayLength = count($result);
for($i = 0; $i < $arrayLength; $i++) {
mysqli_query($con, "DELETE FROM table WHERE id NOT IN ('".$result[$i]['id']."')");
}
If anyone knows what is going on i'd appretiate the help or any suggestions on how to achieve the same result. I am using php 7 and mysql 5.7 with innodb tables if that helps.
It probably doesn't work because your IN value is something like this
IN('1,2,3,4');
When what you want is this
IN('1','2','3','4')
OR
IN( 1,2,3,4)
To get this with implode include the quotes like this
$in = "'".implode("','", $array)."'";
NOTE whenever directly inputting variables into SQL there is security Implications to consider such as SQLInjection. if the ID's are from a canned source you're probably ok, but I like to sanitize them anyway.
You can't mix array and string.
This works:
mysqli_query($con, "DELETE FROM table WHERE id NOT IN (".implode(',', $ids).")");
I have quite an issue I can not seem to solve. I am trying to get a row count from a select statement.
I should start by saying I have tried most all methods resulting from google searches on this issue.
I am using the result set so I would prefer not to make a second query.
The query uses a prepared select statement which seems to be a main issue if I understand it correctly.
I decided to try a simple approach using PHP's native count() function. Which lead me here because I finally reached the end of the rope on this.
On to the details...within a class of mine, I make the query like this.
// Create database connection
$database = DatabaseFactory::getFactory()->getConnection();
// Set and execute database query
$sql = "SELECT * FROM `service_orders` WHERE `agency_id` = :agency_id $filter ORDER BY $sort $order $per_page";
$query = $database->prepare($sql);
$query->execute($query_array);
// If results
if ($query->rowCount() > 0) {
$results = $query->fetchAll();
self::$order_count = "Count: " . count($results);
return $results;
}
// Default, return false
return false;
Findings
If I perform count($results) like I did above, I get the total rows in the database (Let's say 50).
If I print_r($results), it shows the array with the proper number of entries (Let's say 10) that of course differs from the total rows in the database.
How can these two differ? It's as if the count($results) is misreading the result array.
More Findings
Within my actual php page, I call the class to retrieve the data like this.
$results = OrderModel::getServiceOrders();
echo count($results);
Strangely enough, if I then perform count($results) it gives me the correct reading of the result array (which in my example here would be 10).
I am perplexed by this as the count function is being performed on the exact same array. The only difference is one is called on the array within the class, and the other is called on the array returned from the class.
Does anyone have any ideas on how to solve this or why there is the discrepancy when using count() in this instance?
Thank you all in advance!
James
Additional Info
This is another mind numbing scenario. If I return the count along with the actual results, I can access it on the page with the correct value (10 rows). Yet, if I set it into a session variable, and access it that way, the count is the whole data set (50 rows). How is it even possible these two values are not the same?!
$results = $query->fetchAll();
Session::set("order_count", $total[0]); // Yields 50 (incorrect)
return [
"results"=> $results,
"count"=> $total[0], // Yields 10 (correct)
];
I am trying to input multiple pieces of data through a form and all the data will be separated by (,). I plan to use this data to find the corresponding id for further processing through an sql query.
Below is the code I use.
$key_code = explode(",", $keyword);
//$key_count = count($key_code);
$list = "'". implode("','", $key_code) ."'";
//$row_count = '';
$sql4= "SELECT key_id FROM keyword WHERE key_code IN (".$list.")";
if(!$result4 = mysql_query($sql4, $connect)) {
mysql_close($connect);
$error = true;
}else{
//$i = 0;
while($row = mysql_fetch_array($result4)) {
$keyword_id[] = $row['key_id'];
//$i++;
}
//return $keyword_id;
}
The problem i see is that keyword_id[0] is the only element that contains any data (the data is accurate). Even if I input multiple values through the aforementioned form.
I thought it might be an error in the sql but I echo'ed it and it looks like:
SELECT key_id FROM keyword WHERE key_code IN ('WED','WATER','WASTE')
The values in the brackets are exactly what I inputted.
I even tried to figure out how many rows are being returned by the query and it shows only 1. I assume something is wrong with my query but I cannot figure where.
Any help will be greatly appreciated.
Edit: Alright Solved the problem. Thanks to suggestions made I copied and pasted the $sql_query I had echo'ed on the website into mysql console; which resulted in only 1 row being retrieved. After taking a closer look I realized that there was a whitespace between ' and the second word. I believe the problem starts when I input the key_code as:
WED, WATER, WASTE
Instead inputting it as
WED,WATER,WASTE
fixes the problem. I think I should make it so that it works both ways though.
Anyway, thank you for the help.
I am pretty sure that the query is ok. How many rows do you get with just
SELECT key_id FROM keyword
I think that there is just one line that matches your WHERE.
Check the query directly in the database(with phpmyadmin, or in the mysql console), however this query seems to be working as you may assumed. If it returns only 1 row when you use it directly in the db, then maybe there is only one row in your table wich matches this query.