Search and Display Data from SQL Database

Question

I'm trying to search for some data in a database and display it. This is the html form code I have to enter the search criteria.

<h2> Search </h2>
<form action = "search.php" method = "post" >
  Search for: <input type = "text" name ="find" /> in
  <select NAME = "field">
    <Option VALUE = "Animal Type"> Animal Type</option>
    <Option VALUE = "latitude"> Latitude</option>
    <Option VALUE = "longitude"> longitude</option>
    <Option VALUE = "dateseen"> Date Required</option>
    <Option VALUE = "timeseen"> Time</option>
  </select>
  <inpput type= "hidden" name = "searching" value ="yes"/>
  <inpput type= "submit" name = "search" value ="Search"/>
</form>

This is the php code I am using. But I keep gettin an error saying undefined variable at line 18/22 and

mysql_fetch_array() expects parameter 1 to be resource

errors. Any ideas?

<?php

if ($searching=="yes")
  {echo "<h2> Results</h2><p>";
  }

if ($find=="")

{echo "<p> Please enter a search iten";
exit;

}

$link=mysqli_connect("localhost","root","");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$db_selected = mysqli_select_db($link,"Animal_Tracker");

if (!$db_selected)
  {
  die ("Can\'t use test_db : " . mysqli_error());
  }

$find = strtoupper($find);
$find = strip_tags($find);
$find = trim($find);

$sql=mysql_query("Select * FROM Animal_Tracker WHERE upper($field) LIKE '%$find%' ");

while($result = mysql_fetch_array($sql))
{
    echo $result ['Animal Type'];
    echo " ";
echo $result ['latitude'];
echo "<br> ";
echo $result ['longitude'];
echo " <br>";
echo $result ['dateseen'];
echo " <br> ";
echo $result ['timeseen'];
echo "<br> ";
echo "<br> ";
}

You're using a variable called `$field` in your SQL query, but you don't define it anywhere. — David, Feb 15 '14 at 14:30
you are making a mysqli connection and executing mysql_* functions.. — Nouphal.M, Feb 15 '14 at 14:30

score 1 · Accepted Answer · answered Feb 15 '14 at 15:26

I managed to get your code working. You mix both mysql and mysqli (I'm not sure that was the problem though).

Good practice is to check if your mysql_query was successful and print info. Don't forget to use your own field names (I created a test database)

<?php


$link=mysqli_connect("localhost","root","");

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$db_selected = mysqli_select_db( $link, "Animal_Tracker");


 if (!$db_selected){
        die("Couldn't select database $db ".mysqli_error($link));
}


$sql=mysqli_query($link, "Select *  FROM location");

if ($sql == FALSE)
{
  die($sql." Error on query: ".mysqli_error($link)); 
}

while($result = mysqli_fetch_array($sql))
{
    echo $result ['x'];
    echo " ";
    echo "<br> ";
}

Thanks this did help but could you tell me how to use the first six lines of the code as well. How I could define $find and $searching so that it can also the access the HTML code? Otherwise it displays all the data on the database even if I don't enter any information — user3293736, Feb 15 '14 at 20:06
Sure, have a look here http://stackoverflow.com/questions/17139501/using-post-to-get-select-option-value-from-html. It is a very common problem to work with. — VonSchnauzer, Feb 15 '14 at 20:40

Search and Display Data from SQL Database

1 Answers1