How can I get a specific data using SQL select query? - php

Question

I am trying to make a blog site.For this purpose I need to use a specific data from a specific field from my database table.To do that I wrote these code.
<?php
$host = "localhost";
$user = "root";
$pass = "12345";
$db = "bnsb";
$conn = mysql_connect($host, $user, $pass) or die("Connection Failed!");
mysql_select_db($db, $conn) or die("Database couldn't select!");
$img = "select image from news where uid=1";
echo $img;
?>
My database connection is OK.It should print like this user_img1.jpg. But it prints the whole sql query like select image from news where uid=1. I run this code on phpmyadmin. It works! But it does not work in my php script.How can I do now?

Answer 1

You can not give the query as it is and expect result like in phpadmin.
For this first of all you have to connect to your DB like this
$con = mysqli_connect("localhost","my_user","my_password","my_db");
execute required query like this
$query22 = "select image from news where uid = 1";
$result22 = mysqli_query($con, $query22) or die (mysqli_error());
Get the result and display like this
while($rows = mysqli_fetch_array($result22, MYSQLI_BOTH))
{
echo "<br>Values in db: " . $rows['columnname'];
}
Also i advice you to take a look at these tutorials
http://codular.com/php-mysqli
http://www.dreamincode.net/forums/topic/54239-introduction-to-mysqli-and-prepared-statements/

Answer 2

Please read some PHP 101 kind of tutorials on how to use PHP.
To get data from DB (in almost any language)
You need to connect to a DB. The connection gets you some sort of resource
You formulate your query (which you seem to have done)
You execute the query against the DB that you connected to (step #1)
You get a result (set)
You iterate over the result set to get the individual result(s); in your case the result set would be just one result (or row).
The examples to do this in PHP are very basic; please do your own lookup on net. This one seems good enough to get you started - http://www.w3schools.com/php/php_mysql_intro.asp

Answer 3

Try this,
<?php
$host = "localhost";
$user = "root";
$pass = "12345";
$db = "bnsb";
$conn = mysql_connect($host, $user, $pass) or die("Connection Failed!");
mysql_select_db($db, $conn) or die("Database couldn't select!");
$img = "select image from news where uid=1";
$result=mysql_query($img);
while($row=mysql_fetch_array($result)){
echo '<img src="your_path_to_image/'.$row['image'].'" /> - '.$row['image'];
}
?>