PHP variable in $row[]

Question

Any idea how to include variable in $row[] mysqli_query?

what I've tried until now:

1

$result = mysqli_query($connection, "SELECT ".$var." AS Variables, COUNT(*)  FROM data GROUP BY '$var'"); 
while($row = mysqli_fetch_array($result)) {
    echo $row['Variables'];
    echo $row['COUNT(*)']; 
}

2

$result = mysqli_query($connection, "SELECT '$var', COUNT(*)  FROM data GROUP BY '$var'"); 
while($row = mysqli_fetch_array($result)) {
    echo $row[$var];
    echo $row['COUNT(*)']; 
}

Also, I've tried all combination

echo $row[$var];
echo $row["$var"];
echo $row['$var'];

I really REALLY hope $var is not submitted by the user... — Federico klez Culloca
– Federico klez Culloca, Commented Apr 3, 2018 at 13:53
Please explain a bit more. Including a vaiable in $row is what you have been doing in $row[$var] — Ashish Choudhary
– Ashish Choudhary, Commented Apr 3, 2018 at 13:54
Anyway, the first one is good (except for the risk of injection), just group by Variables instead and add an alias for the count, so you can use that as an index for $row. — Federico klez Culloca
– Federico klez Culloca, Commented Apr 3, 2018 at 13:57
both codes I wrote is not working, group by Variables not returning anything — Serghei Pogor
– Serghei Pogor, Commented Apr 3, 2018 at 14:14
Don't build SQL queries by string concatenation. It is error-prone and vulnerable to SQL-injection. Use prepared statements. — axiac
– axiac, Commented Apr 3, 2018 at 14:25

Grant · Accepted Answer · 2018-04-03 14:11:19Z

2

Try this:

$result = mysqli_query($connection, "SELECT ".$var." AS Variables, 
COUNT(*) AS count FROM data GROUP BY '$var'"); 
while($row = mysqli_fetch_array($result)) {
    echo $row['Variables'];
    echo $row['count']; 
}

answered Apr 3, 2018 at 14:11

Grant

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4 Comments

Serghei Pogor Over a year ago

thanks, it's work now. So, the issue was with COUNT(*)... didn't think about it at all

Grant Over a year ago

Just create an alias instead of selecting just COUNT(*), then its easy to access that in the while loop

Grant Over a year ago

NB: You have a GAPING SECURITY HOLE in your SQL statement, you need to ensure that a user cannot inject code into your SQL query by manipulating the $var variable. SQL Injection: acunetix.com/websitesecurity/sql-injection Use PDO or prepared statements instead.

Serghei Pogor Over a year ago

thanks for your advice, it's for a local server, no one has access except admin

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