I want it to echo how many posts there are in bestallt where its ID is 1 from the table order and display it as how many posts there are in numbers. I am connected to the database in the PHP file, that's not an issue. I'm just not sure how to put it all down in PHP, quite new to this. I just can't get it to echo what I want, nothing comes out/I get an error.
Any help is appreciated!
Your present code does not fetch the data from the database, it simply echoes the SQL query you have written.
Quite a lot more code goes in to fetching results from a mysql database, and I am not sure reproducing a full explanation here will serve anyone's interests. However, you may wish to view the examples of how to use PDO (a method of using mysql databases from php) here, and then either edit or re-ask your question if you find you have specific difficulties following those examples.
Try this code:
<td>
<?php
$sql = "SELECT count(*) FROM order WHERE bestallt='1'"; // Your SQL query
$response = mysql_query($sql);
if (mysql_num_rows($response) == 0) {
// Your query returned 0 rows!
}
while($row = mysql_fetch_array($response)){
// For each row returned from your query
// $row is an array
// For example, You can use it:
echo $row['data1'];
echo $row['data2'];
echo $row['data3'];
// data1, data2, data3 are the name of the fiels in your database
}
?>
</td>
Have a nice day!
Try this,
$sql = "SELECT count(*) FROM order WHERE bestallt='1'"
$res=mysql_query($sql);
$arr=mysql_fetch_array($res);
echo "<pre>";print_r($arr); //you will get whole array for matching record of your order table if found anything
To learn more, you can check my earlier Answer: how generate report between the two dates using datepicker,ajax,php,mysql.?
In that answer I have described whole working demo.
Related
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 = $row_bolts['name'];
} //bolts while
} // bolts num rows
I am not using mysql, but mysqli. I did a whole bunch of tutorials on PDO and how to connect to the database and retrieve information and just couldn't figure out how to show values from a row like I can with the above. It's frustrating because I want to use the latest methods, but I can't find how to online that makes sense to me.
Thank you for eveyone's comments so far.
This question already has answers here:
fetch single record using PHP PDO and return result
(2 answers)
Closed 6 years ago.
I have not made the change from MySQL_ to PDO until today. Needless to say, the migration is more than a simple headache. So, I need a bit of help. I tried all the search terms I could before registering and asking this question.
My Problem
User types a numeric code into the search box, translates it to
.php?code=term
Script selects all columns from the database where the code is the
code term searched for.
PHP will Echo the results
My Code
if (isset($_GET["code"])) {
//IF USER SEARCHES FOR CODE, RUN THIS. ELSE SKIP.
$crimecode = $_GET["code"];
$crcode = $link->prepare("SELECT * FROM crimecodes WHERE code = :code");
$crcode->bindParam(':code', $crimecode);
$crcode->execute();
$coderesult = $crcode->fetchAll();
echo "<h4>CODE:</h4>";
echo $crimecode;
echo "<br /><h4>DEFINITION:</h4>";
echo $coderesult;
die();
}
Before, it was simple. All I had to do was:
$qcode = mysql_query("SELECT * FROM crimecodes WHERE code = $crimecode");
$fcode = mysql_fetch_assoc($qcode);
echo $fcode['definition'];
But, the ever evolving world has decided to fix something that wasn't broken so now the whole prior code is pointless and you gotta learn something new. Any help is appreciated to get this to work.
Right now, the above PDO code returns definition: ARRAY.
Like literally, the $coderesult prints Array.
The fetchAll() option returns an array containing all of the result set rows (http://php.net/manual/pt_BR/pdostatement.fetchall.php).
$coderesult prints Array because it's actually an array. If you do var_dump($coderesult) you'll see it.
I suppose that you are trying to get one row only. If that's the case, add this line after $coderesult = $crcode->fetchAll();:
$coderesult = $coderesult[0];
Then you can
echo $coderesult['definition'];
If you're trying to get more than one row, you need to use foreach to loop through the array.
I suggest you read the php manual for PDO Class or mysqli, wherever you prefer. There's a lot more options than mysql_.
Also, I think it's worth to mention that your previous code
$qcode = mysql_query("SELECT * FROM crimecodes WHERE code = $crimecode");
$fcode = mysql_fetch_assoc($qcode);
echo $fcode['definition'];
it's vulnerable to SQL Injection.
Hi there i am working on PHP code that is selecting columns from two tables.
Here is my code:
$result2 = mysql_query("SELECT *
FROM `videos`, `m_subedvids`
WHERE `videos.approved`='yes' AND
`videos.user_id`='$subedFOR'
ORDER BY `videos.indexer`
DESC LIMIT $newVID");
while($row2 = mysql_fetch_array($result2))
{
$indexer = addslashes($row2['videos.indexer']);
$title_seo = addslashes($row2['videos.title_seo']);
$video_id = addslashes($row2['videos.video_id']);
$title = addslashes($row2['videos.title']);
$number_of_views = addslashes($row2['videos.number_of_views']);
$video_length = addslashes($row2['videos.video_length']);
}
When i try to print $indexer with echo $indexer; it's not giving me any results.
Where is my mistake in this code?
It seems to me like the key 'indexer' isn't in your results. It's hard to tell, since you haven't listed a definition for your table and you're using SELECT * so we can't see the names.
It makes the program easier to read later, if instead of SELECT *..., you use SELECT col1, col2, .... Yes, SELECT * will save you some typing right now, but you'll lose that time later when you or anyone else who works on your code has to check the table definition every time they work with that line of code.
So, try changing your query to explicitly select the columns you use. If it's an invalid column you'll get an error right away rather than this silent failure you're getting now, and you'll thank yourself later as well.
So long as videos.indexer is a unique field name among all tables used in the query you can change
$indexer = addslashes($row2['videos.indexer']);
to
$indexer = addslashes($row2['indexer']);
You don't need to (or can not) use the table name when referring to the result.
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.
Got a relatively simple MySQL query that I'm pulling using php with the following code:
$employeeNames = mysql_query(
"SELECT *
FROM employees
WHERE team=\"1st Level Technical Support_a\"
LIMIT 0,5000") or die(mysql_error());
$employeeNumRows = mysql_num_rows($employeeNames);
echo $employeeNumRows;
while ($row = mysql_fetch_array($employeeNames, $employeeNumRows)) {
echo $row['full_name'];
}
Now, if I run the query on the first line in SQL it gives me 18 results. If I echo $employeeNumRows it prints 18. Nothing else after that though.
If I change "1st Level Technical Support_a" to any other team in the table, it will bring up the proper results using PHP
This is the weirdest problem I've come across using MySQL/PHP - can anyone help? Has anyone seen something like this before?
Try removing the second parameter from your call to mysql_fetch_array, so that it reads mysql_feetch_array($employeeNames). See the documentation of the function to see how to use it properly.