Display data from table with specific column values via php

Question

this ought to be simple but am yet to find an answer for it (i have searched the questions in stackoverflow). am on php and i have a table books on mysql. What i want is a list displayed in my webpage with these specifics on a table created by php. I know the mysql code:

SELECT title FROM books WHERE category='currently reading' 

applying that on php has brought this error, Parse error: syntax error, unexpected 'currently' (T_STRING)

Here is my php code:

<?php
include('include/databaseconnection.php');
include('include/insertingbooks.php');
// selecting data 
$result = mysql_query ('SELECT title FROM books WHERE category='currently reading'';);
//("SELECT title FROM books WHERE category LIKE $currently");
//opening table tag
echo "<table border = 1px>";
while ($data = mysql_fetch_array($result)) {
// printing table row
echo'<tr>';
echo '<td>'.$data['title'].'</td>';
echo'</tr>'; // closing table row
}
echo '</table>';
?>

If i decide to leave out WHERE clause, it works perfectly except it displays all the books. The options i have tried already

• using WHERE category LIKE $category while setting up a variable $category = "currently reading"; but it dint work.
• trying to link it to the form i got the the values of category from by adding include('include/insertingbooks.php'); (which contains $category =$_POST ['category'];) and trying to put $category.
• using WHERE not but it didnt work at all.

Answer 1

You have bad quotes and an extra semi-colon here -

$result = mysql_query ('SELECT title FROM books WHERE category='currently reading'';);

Change to this (note the double quotes) -

$result = mysql_query ("SELECT title FROM books WHERE category='currently reading'");

In addition, you should stop using mysql_* functions. They are no longer maintained and are officially deprecated. Learn about prepared statements instead, and use PDO.

You should also add MySQL error checking to your queries and connections.

Using or die(mysql_error()) to mysql_query().

For example:

$result = mysql_query ("SELECT title 
                        FROM books 
                        WHERE category='currently reading'") 
    or die(mysql_error());

• Which would have caught and displayed the syntax error.

Answer 2

Use this code:

<?php
include('include/databaseconnection.php');
include('include/insertingbooks.php');
// selecting data 
$result = mysql_query ("SELECT title FROM books WHERE category='currently reading'");
//("SELECT title FROM books WHERE category LIKE $currently");
//opening table tag
echo "<table border = 1px>";
while ($data = mysql_fetch_array($result)) {
// printing table row
echo'<tr>';
echo '<td>'.$data['title'].'</td>';
echo'</tr>'; // closing table row
}
echo '</table>';
?>

Answer 3

there is only quotes problem Please replace your query to:

$result = mysql_query ("SELECT title FROM books WHERE category='currently reading'";);

Please note you can not use same quotes eg. If you are using double quotes then use single quotes inside and if single quotes then double code inside