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

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?

you have to execute the query and get the result and then print it.but as of now you are just printing a string — krishna
– krishna, Commented Feb 11, 2014 at 6:44
connect to your db... select your db... run the $img query... get the value of the result... and you are happy — silly
– silly, Commented Feb 11, 2014 at 6:45
Heres a good starting point for you, showing how to use mysqli_* functions in php, from connecting to the database to executing your query and dealing with the result set; pontikis.net/blog/… - hope this helps :) — flauntster
– flauntster, Commented Feb 11, 2014 at 6:47
@user3167680 again you are trying to print only the query without executing it.Please execute it and get the result and print that result — krishna
– krishna, Commented Feb 11, 2014 at 7:04

krishna · Accepted Answer · 2014-02-11 07:08:31Z

3

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/

edited Feb 11, 2014 at 7:08

answered Feb 11, 2014 at 6:51

krishna

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8 Comments

Kallol Das Over a year ago

Thanks krishna. I have tried like you said. but there was an error message. "Query was empty". @krishna

krishna Over a year ago

can you post the code which you tried ? Mainly the querys and execution part

Kallol Das Over a year ago

<?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 sn=1"; $result=mysql_query($conn,$img); $row=mysql_fetch_array($result); echo $row['image']; if(mysql_query($result)) { echo "congratulations!"; } else { die(mysql_error()); } ?> The result is "Query was empty" @krishna

krishna Over a year ago

the problem is you are mixing mysqli syntax with mysql syntax. change mysql_query($conn,$img); to mysql_query($img,$conn);

krishna Over a year ago

i also advice you to use mysqli_* functions

|

Amit · Accepted Answer · 2014-02-11 06:46:07Z

1

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

answered Feb 11, 2014 at 6:46

Amit

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1 Comment

Kallol Das Over a year ago

Thanks. I m trying for two days in various websites and books included(w3schools). But have not got any solution yet.So I m asking here.For your kind information, My database connection is OK.Thanks again.:-)

Sherin Jose · Accepted Answer · 2014-02-11 07:02:15Z

0

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'];
}
?>

edited Feb 11, 2014 at 7:02

answered Feb 11, 2014 at 6:49

Sherin Jose

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