php/sql- How do I use LIKE operator on variable to pull results from database?

Question

As question said, I want to use the LIKE Operator so whenever user inputs something, like "M", it returns all database entries starting with "m" rather then anything named just "M". This is my code

$strSymbolName = @$_REQUEST["searchSymbol"];
//searchSymbol is the variable with user Input inside

if(!empty($strSymbolName))
{ // process the form
 // Establish dbserver connection and default database

$db = $objDBUtil->Open();
 // Run a Query to get company name

$query = "SELECT symSymbol, symName FROM symbols " .
 "WHERE symSymbol =" . $objDBUtil->DBQuotes($strSymbolName);
//Retrieves company symbol and name from database. Stores this in result

$result = $db->query($query);`

What I thought of doing was using the LIKE operator on the WHERE statement, so something like-

WHERE symSymbol LIKE = " . $objDBUtil->DBQuotes($strSymbolName);

But this didn't work...How would I apply the LIKE operator here?

Get rid of the = sign after LIKE and then use % wildcard markers where appropriate. — Mike
– Mike, Commented Jun 8, 2017 at 2:40
I tried that line, replaced it with what I showed above, but my program isn't recognizing it (the line is gray) "WHERE symSymbol "LIKE %".$objDBUtil->DBQuotes($strSymbolName)."%" — Haha Haha
– Haha Haha, Commented Jun 8, 2017 at 3:30

dylan · Accepted Answer · 2017-06-08 03:40:16Z

1

$db = $objDBUtil->Open();

$param = $objDBUtil->DBQuotes($strSymbolName);

$query = "SELECT symSymbol, symName FROM symbols WHERE symSymbol LIKE '%$param%'";

$result = $db->query($query);

edited Jun 8, 2017 at 3:40

answered Jun 8, 2017 at 2:44

dylan

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3 Comments

Syfer Over a year ago

Welcome to SO. Can you also put in some explanation of what the code does and format it correctly please?

dylan Over a year ago

$db = $objDBUtil->Open(); $param = $objDBUtil->DBQuotes($strSymbolName); $query = "SELECT symSymbol, symName FROM symbols WHERE symSymbol LIKE '%$param%' "; $result = $db->query($query);

Syfer Over a year ago

I meant in your answer. Format the code using the code tags and put in some explanation for users as well please

alil rodriguez · Accepted Answer · 2017-06-08 06:10:57Z

1

just use LIKE query

$db = $objDBUtil->Open();
$query = "SELECT symSymbol, symName FROM symbols WHERE symSymbol LIKE '%$strSymbolName%'";
$result = $db->query($query);

answered Jun 8, 2017 at 6:10

alil rodriguez

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

If you want to search something that starts with "M", add a % after the value, and remove the = as it is not needed when using LIKE

$query = "SELECT symSymbol, symName FROM symbols WHERE symSymbol LIKE ".$objDBUtil->DBQuotes($strSymbolName)."%";

For more information about LIKE, check here.

WHERE CustomerName LIKE 'a%'        || Finds any values that starts with "a"

WHERE CustomerName LIKE '%a'        || Finds any values that ends with "a"

WHERE CustomerName LIKE '%or%'    || Finds any values that have "or" in any position

WHERE CustomerName LIKE '_r%'       || Finds any values that have "r" in the second position

WHERE CustomerName LIKE 'a_%_%' || Finds any values that starts with "a" and are at least 3 characters in length

WHERE ContactName LIKE 'a%o'          || Finds any values that starts with "a" and ends with "o"

Brian Moreno · Accepted Answer · 2017-06-08 02:45:50Z

Use the % wildcard after the string you want to look for.

$query = "SELECT symSymbol, symName FROM symbols WHERE symSymbol LIKE ".$objDBUtil->DBQuotes($strSymbolName)."%";

A bit more on the % and other wildcards: https://www.w3schools.com/sql/sql_wildcards.asp

Excerpt from above link:

WHERE CustomerName LIKE 'a%' || Finds any values that starts with "a"

WHERE CustomerName LIKE '%a' || Finds any values that ends with "a"

WHERE CustomerName LIKE '%or%' || Finds any values that have "or" in any position

WHERE CustomerName LIKE '_r%' || Finds any values that have "r" in the second position

WHERE CustomerName LIKE 'a_%_%' || Finds any values that starts with "a" and are at least 3 characters in length

WHERE ContactName LIKE 'a%o' || Finds any values that starts with "a" and ends with "o"

Sam · Accepted Answer · 2017-06-08 02:53:53Z

0

You need to append the % on the end of your variable like:

WHERE symSymbol LIKE " . $objDBUtil->DBQuotes($strSymbolName) . "%";

This will return anything starting with your variable.

edited Jun 8, 2017 at 2:53

answered Jun 8, 2017 at 2:39

Sam

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