PHP Parse Error

[deslyxia]

I am getting the following code when executing a query:

 

Parse error: syntax error, unexpected T_VARIABLE

 

 

Below is the query it came from.

 

//query db and build name array
$qry="SELECT * FROM Patient WHERE LName LIKE "$q"";
$result=mysql_query($qry);

 

I know that the issue is coming from the double quotes around $q. If I switch them to single quotes I get no error... but I also get no results. I have looked up the difference between single and double quotes and I think i have it right.

 

Single quotes use exactly what is in them  ex.  echo '$q';  would be $q

Double quotes will evaluate the variable and use that 

ex.  $q = 'php';

echo "$q";  would be php

 

So this has me kind of lost as to what the issue is in my query.

[PFMaBiSmAd]

The correct php and mysql syntax would be -

 

$qry="SELECT * FROM Patient WHERE LName LIKE '$q'";

 

Using LIKE without any wildcard characters in the pattern is in most cases the same as using an equal comparison -

 

$qry="SELECT * FROM Patient WHERE LName = '$q'";

 

What LName value are you trying to match and what value are you putting into the $q variable?

[deslyxia]

In my Table I have hundreds of LName. $q comes from a user input field on the main page. The purpose here is that every time a user types a letter in the field this query selects all rows from my table that contain the user input string.

[PFMaBiSmAd]

I'll assume you mean you want to find names that start with the entered string. You would use a trailing % wildcard to match everything that starts with the entered string -

 

$qry="SELECT * FROM Patient WHERE LName LIKE '$q%'";