Compare with database values and find search result in PHP

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passing two values in url treatment and city from home.php

and in show.php

I want to display data according to these two values if found

in data base city column goes like Delhi(single value) but treatment column goes like prp, fever, infection(multiple values in single cell)

my problem is how to select and display if url consists treatment = prp and city = mumbai
then display no data found because city has delhi

What I have tried:

$sql = "SELECT * FROM add_doctor WHERE city='".$city."' AND treatment ='".$treatment."'";

Posted 11-May-17 22:02pm

Ajay Singh

Updated 12-May-17 3:20am

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Richard Deeming 12-May-17 8:03am

So your problem is that there are no matching records in the database, and you don't know why it's not finding any data to display?

Also, storing multiple values in a single field is a very bad database design. You should normalize your database, and have the multiple treatment values in a separate table with a foreign-key link to the doctors table.

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W∴ Balboos, *GHB* · Answer 1 · 2017-05-12T03:20:00

Two changes to make.

1) Since you're using .php, you can save youself some trouble when using " as your string wrapper.

$sql = "SELECT * FROM add_doctor WHERE city='".$city."' AND treatment ='".$treatment."'";
Could be written: 
$sql = "SELECT * FROM add_doctor WHERE city='$city' AND treatment ='$treatment'";

The php parser will replace the $ variable at run-time.

Now, to your query. Rewrite to something like:

$sql = "SELECT * FROM add_doctor WHERE city='$city' AND treatment LIKE '%$treatment%'";

With the '%' wild-cards wrapping your diseases. This will match any string that contains the value in $treatment regardless of what surrounds it.

As noted in first comment, the design for the disease field is very bad.