how do i display a field that i have got from the address bar?

Question

how do I display a field when I have used a GET command to get it from the address bar from the previous page and then display it in my table?

I have tried to display the field already in the table but no luck. the field is call "Reference"

using the below I want to get the reference into the table. (I have tried the following echo "" . $row['$Reference'] . "";)

http://test.com/searchresults.php?Reference=456789 $Reference=$_GET['Reference'];

search results page

<?php

require_once('auth.php');

$host=""; // Host name 
$username=""; // Mysql username
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="Physio"; // Table name 

 // Connect to server and select database.
   mysql_connect($host, $username, $password)or die("cannot connect"); 
   mysql_select_db($db_name)or die("cannot select DB");

   // get value of Reference that sent from address bar
 $Reference=$_GET['Reference'];

  $sql="SELECT * FROM $tbl_name WHERE Reference='$Reference'";
 $result=mysql_query($sql);

 $rows=mysql_fetch_array($result);

    if(!isset($_POST['postcode'])) {
  header ("location:index.php");
 } 
     echo "<p> Results </p>" ;
 $search_sql = "SELECT * FROM `Physio` WHERE Postcode like '%" . $_POST['postcode'] . "%'";
$search_query = mysql_query($search_sql);

while ($search_rs = mysql_fetch_array($search_query))
{
echo"<table><tr>" ;
     echo '<th>Reference</th><th>Select</th><th>PhysioReference</th><th>Physio</th><th>Postcode</th><th>Line1</th><th>Tel</th>';
while ($search_rs = mysql_fetch_array($search_query))
     {
     echo '<tr>';
     echo "<td>" . $row['$Reference'] . "</td>";
     echo "<td><a href=\"Physiotoinstruction.php?Reference={$search_rs['Reference']}\">Select</a></td>";
     echo "<td>".$search_rs['PhysioReference'] . "</td>";
     echo "<td>".$search_rs['Name'] . "</td>";
     echo "<td>".$search_rs['Postcode'] . "</td>";
     echo "<td>".$search_rs['Line1'] . "</td>";
     echo "<td>".$search_rs['Tel'] . "</td>";
     echo '</tr>';
     }
  echo '</table>';
}
if (!empty($search_rs))
{
echo "<p> Results Found </p>" ;
}
else
{
    echo "<p> No Results Found </p>" ;
}

   ?>

Please have a look at http://bobby-tables.com/ – lethal-guitar Mar 25 '14 at 10:28 — lethal-guitar, Mar 25 '14 at 10:28

GMemory · Accepted Answer · 2014-03-25T10:50:12.317

2

Variables will not expand when in single quotation marks, and also you're trying to access variable called $row which isn't in scope (i.e. doesn't exist at this point). The following should suffice:

$rows[$Reference]

Or change $rows to $row earlier in your code, whichever makes most sense semantically.

edited Mar 25 '14 at 10:50

answered Mar 25 '14 at 10:22

GMemory

137
9

same as other answer, i have tried these and they seem correct but still no data is showing, have i used the GET command correctly to enable the display the reference? – user3301611 Mar 25 '14 at 10:34
$row hasn't been initialized to anything at that point. $rows has, but there's no variable called $row before you try to access it. I'll update my answer to include this. – GMemory Mar 25 '14 at 10:46
You appear to be using $rows instead of $row. So either change $rows to $row or vice versa (considering you only appear to be fetching one result, $row would probably make the most sense) – GMemory Mar 25 '14 at 10:51
I might have used the wrong code here looking at it,i think i need to remove the rows. - how do i display the reference, from the $_GET['Reference']; and display it in the column of the table. many thanks in advance – user3301611 Mar 25 '14 at 11:00
`echo "" . $_GET['Reference'] . "";` or `echo "" . $Reference . "";` if you keep the assigned variable. – GMemory Mar 25 '14 at 11:05
quick question, how would I put 2 variables values into url. the following plus the reference I just put into this table "reference" --- echo "Select"; – user3301611 Mar 25 '14 at 11:16
GET parameters are delimited by the `&` character (and remember that `&` is a special character in HTML, so will need to be encoded as `&`) - `echo "Select";`. If you have any further questions, you might want to consider opening a new question as we're going well beyond the scope of the original question here. – GMemory Mar 25 '14 at 11:20

score 1 · Answer 2 · answered Mar 25 '14 at 10:23

1

Instead of $row['$Reference'] use $row[$Reference].

Apart from that DO NOT USE mysql_* SINCE IT IS DEPRECATED. Use mysqli or PDO.

Also, escape parameters which you receive from the web. Otherwise you are vulnerable to mysql injection attacks.

answered Mar 25 '14 at 10:23

dkasipovic

5,930
1
19
25

+1 for the SQL injection warning. Still way too much bad PHP tutorials out there.. – lethal-guitar Mar 25 '14 at 10:26
i have tried these and they seem correct but still no data is showing, have i used the GET command correctly to enable the display the reference? – user3301611 Mar 25 '14 at 10:32
You should show your table structure and example of data in it if you need more help. – dkasipovic Mar 25 '14 at 10:33

score 0 · Answer 3 · edited May 23 '17 at 12:05

0

You can use following methods

$row["$Reference"] // with double quotes

or

$row[$Reference]  //no quotes

As with single quotes it is considered as string.

Read this for reference.

edited May 23 '17 at 12:05

Community

1
1

answered Mar 25 '14 at 10:24

عثمان غني

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how do i display a field that i have got from the address bar?

3 Answers3