如何使用SQL选择查询获取特定数据？

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)

1. You need to connect to a DB. The connection gets you some sort of resource
2. You formulate your query (which you seem to have done)
3. You execute the query against the DB that you connected to (step #1)
4. You get a result (set)
5. 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'];
}
?>