Data fetched from mysql is displayed 2 times

Question

I am new to php. I want to fetch a particular data from mysql and display it in the label.I have tried a simple php coding.But it displays the fetched data two times(actually I have created 2 columns such as name and age in a table called test).Please help me.Here is the coding:

displayform.php

<body>
 <form method="post" name="display" action="display.php" >
 Enter the name you like to display the data from MySQL:<br>
<input type="text" name="name" />
<input type="submit" name="Submit" value="display" /> </form> 
</body>
</html>

display.php

<?php 
mysql_connect("localhost", "root", "") or die("Connection Failed");
 mysql_select_db("acp")or die("Connection Failed"); 
 $name = $_POST['name']; 
 $query = "select age from test where name = '$name'";
  $result = mysql_query($query); 
  while ($line = mysql_fetch_array($result)) 
  { 
  echo $line['age'];
   echo "<br>\n"; 
   } 
?>

The datas in table is

name=janani
age=25

The output is displayed as

25
25

Are you sure that there is only one `janani` in `test` table? — Francois, Oct 15 '14 at 09:29
Are you sure you don't have duplicated data? If so, add `LIMIT 1` and/or `DISTINCT` to your query. — Funk Forty Niner, Oct 15 '14 at 09:29
use echo mysql_num_rows($result); and find the result num of rows..As a stater you should not blindly depend on result..follow the procedure and see num of rows you are getting — Choco, Oct 15 '14 at 09:30
You might have **janani** name two times.Because code is working perfecty fine.Check your database. — Rohil_PHPBeginner, Oct 15 '14 at 09:34

score 1 · Answer 1 · edited May 23 '17 at 12:27

I am certain that you have two rows bearing the same name and/or age.

In order to show only one result, what you need to do is:

Use either DISTINCT with GROUP BY, or LIMIT 1.

I.e.:

$query = "select DISTINCT age from test where name = '$name' GROUP BY name";

or

$query = "select age from test where name = '$name' LIMIT 1";

Sidenote: I suggest you use mysqli with prepared statements though, since your code is open to SQL injection.

<?php

$DB_HOST = "xxx"; // Replace
$DB_NAME = "xxx"; // with
$DB_USER = "xxx"; // your
$DB_PASS = "xxx"; // credentials

$conn = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($conn->connect_errno > 0) {
  die('Connection failed [' . $conn->connect_error . ']');
}

if($statement=$conn->prepare("select age from test where name = ? LIMIT 1")){

// or this one
// if($statement=$conn->prepare("select distinct age from test where name = ? group by name")){

$name = "janani";

 $statement-> bind_param("s", $name);

// Execute
 $statement-> execute();

// Bind results
 $statement-> bind_result($age);

// Fetch value
 while ( $statement-> fetch() ) {
 echo $age . "<br>";
 }

// Close statement
 $statement-> close();
 }

// Close entire connection
 $conn-> close();

score 0 · Answer 2 · answered Oct 15 '14 at 09:28

0

$line = mysql_fetch_array($result) ; //remove while loop

echo $line['age'][0];

try this ans this is work for one data fetch from table .. And may possible your result show two time because $name match two times in table so it fetch two record

answered Oct 15 '14 at 09:28

Affan

1,132
7
16

This only works if he has a single row in the result. What if he is trying to display multiple lines (aside the duplication he reports, which should be fixed)? – Simone Oct 15 '14 at 09:35
@simone yup I mention that .i think user query is right otherwise – Affan Oct 15 '14 at 09:36

score 0 · Answer 3 · answered Oct 15 '14 at 09:32

0

You are hard coding 'age' in the array reference so it will echo that element only. Loop over array by index and you will get the name also.

answered Oct 15 '14 at 09:32

Crowcoder

11,250
3
36
45

1

This doesn't explain why the data is shown two times. – Simone Oct 15 '14 at 09:34

Data fetched from mysql is displayed 2 times

3 Answers3