php - search form with drop down - php

I am trying to create a basic advanced search with an input which will then search through any results that have a matching category field that is selected in the dropdown and then also a matching keyword field for company_name in "advancedSearch". I have gotten to the stage where I can use the drop down to then display the matching data but I’m having trouble querying that with the search input.
Here is my form code from index.php
<form action="advanced-search.php" method="POST">
<input id="advancedInput" placeholder="Advanced Search" type="search" name="advancedSearch">
<?php
$sqlSelect="SELECT category FROM categories";
$result = $db -> query ($sqlSelect);
echo "<select id=\"selectAdvanced\" name=\"value\">";
echo "<option></option>";
while ($row = mysqli_fetch_array($result)) {
$rows[] = $row;
}
foreach ($rows as $row) {
print "<option value='" . $row['category'] . "'>" . $row['category'] . "</option>";
}
echo "</select>";
?>
<input type="submit" value="search"/>
</form>
And here is the code from my advanced-search.php
<?php
if(isset($_POST['value']) && !empty($_POST['value'])) {
$username = trim(strip_tags($_POST['value']));
include('dbConfig.php');
if (mysqli_connect_errno()) {
printf("Can't connect: %s\n", mysqli_connect_error());
exit();
}
$where = ($username == "category")? "" : " WHERE category = '$username'";
$sql = "SELECT * FROM company_listings" . $where; // Create Query
$result = mysqli_query($db, $sql); // Run Query
echo "<table border=1><tr><th>id</th><th>name</th><th>created</th></tr>";
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['company_name'] . "</td>";
echo "<td>" . $row['created'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
}
This code works great for echoing out the matching categories from the dropdown but I cant work out how I would further query the search from the "advanced search" input.
Any help would be greatly appreciated.

I worked out that using AND in my query like this and using the Request on submit I was able to get what I was after... Thanks to #ADyson for the nudge. I have added my updated advanced-search.php file below;
if(isset($_REQUEST['submit'])){
$username = (($_POST['value']));
$advanced = (($_POST['advancedSearch']));
include('dbConfig.php');
if (mysqli_connect_errno()) {
printf("Can't connect: %s\n", mysqli_connect_error());
exit();
}
$sql=" SELECT * FROM company_listings WHERE company_name like '%".$advanced."%' AND category LIKE '%".$username."%'";
$result = mysqli_query($db, $sql); // Run Query
echo "<table border=1><tr><th>id</th><th>name</th><th>created</th></tr>";
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['company_name'] . "</td>";
echo "<td>" . $row['created'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
}

Related

How I can display all rows in php

I wrote this code to retrieve some rows form database
session_start();
$con = mysqli_connect('localhost', 'root', '');
if(!$con)
{
die("not ok");
}
mysqli_select_db($con,"uoh");
$q = " SELECT * FROM student WHERE id = " . $_SESSION['user_id'] ." and password = " . $_SESSION['user_pass'];
$result = mysqli_query($con , $q ) ;
if($row = mysqli_fetch_array($result))
{
echo "this academic transcripts for " . $row["name"];
echo " and the id is " . $row["id"];
}
$q1 = " SELECT student_record.course,student_record.grade,student_record.term,coe_courses.crd
FROM student_record INNER JOIN coe_courses ON student_record.course_number = coe_courses.course_number
where student_record.id = ".$_SESSION['user_id'] ;
$result = mysqli_query($con , $q1 ) ;
if($row = mysqli_fetch_array($result))
{
echo "<br />";
echo "<table border=\"1\" style=\"width:500\">";
echo "<tr>";
echo "<th>coe_courses</th>";
echo "<th>terms</th>";
echo "<th>Grades</th>";
echo "<th>CRD</th>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row["course"]. "</td>";
echo "<td>" . $row["term"]. "</td>";
echo "<td>" . $row["grade"]. "</td>";
echo "<td>" . $row["crd"]. "</td>";
echo "</tr>";
echo "</table>";
}
The problem is that only shows the first row while I have three rows in phpMyAdmin.
enter image description here
You need to call fetch_* repeatedly to retrieve all rows from your result set; each time you call it it retrieves the next row in the result set.
In your sample code above, you would replace
if ($row = mysqli_fetch_array($result))
{
with
while ($row = mysqli_fetch_array($result))
{
This will loop until fetch_array tries to read beyond the last record in $result, at which point fetch_array returns false and the loop exits.

php mysql table creation on web browser need to add in headings

//replace the following with your details. Dbname is your username by default.
$con = mysqli_connect("info20003db.eng.unimelb.edu.au","geehwank","2058","geehwank");
// Check connection
if (mysqli_connect_errno()) {
echo "Could not connect to MySQL for the following reason: " . mysqli_connect_error();
}
/* this lists the name and release date of all action movies */
echo "<table border='1'>";
$result = mysqli_query($con,"SELECT name, release_date FROM Movie WHERE genre = 'action'");
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['name'] . "</td><td>" . $row['release_date'] . "</td>";
echo "</tr>";
}
mysqli_close($con);
This code is from my uni
and it reads in tables from mysql
and displays them in a table
im a php noob
ive been trying to add headnings to the table called
name, releasedate
how can i do this in code??
can anyone help??
Here you go:
echo "<table border='1'><thead><tr><th>Name</th><th>Release Date</th></tr></thead><tbody>";
$result = mysqli_query($con,"SELECT name, release_date FROM Movie WHERE genre = 'action'");
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['name'] . "</td><td>" . $row['release_date'] . "</td>";
echo "</tr>";}
mysqli_close($con);
?>
</tbody></table>
Try as below :
<html>
<head><title>Lab F</title></head>
<body>
<h1> Lab F - SELECT Example</h1>
<p>
This PHP code runs an SQL query, and displays the answers in an HTML table.
</p>
<p>
<br>
<?php
//replace the following with your details. Dbname is your username by default.
$con = mysqli_connect("info20003db.eng.unimelb.edu.au","geehwank","2058","geehwank");
// Check connection
if (mysqli_connect_errno()) {
echo "Could not connect to MySQL for the following reason: " . mysqli_connect_error();
}
/* this lists the name and release date of all action movies */
echo "<table border='1'>";
$result = mysqli_query($con,"SELECT name, release_date FROM Movie WHERE genre = 'action'");
echo "<tr><td>Name</td><td>Release Date</td></tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['name'] . "</td><td>" . $row['release_date'] . "</td>";
echo "</tr>";}
mysqli_close($con);
?>
</table>
</body>
</html>

PHP mysql query returns a blank screen

I've commented out the bottom part, and the SQL query works fine. Its the displaying of the query where the error is coming from i believe.
$host = "127.0.0.1";
$user = "root";
$pass = "Toom13371!";
$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
// 2. Selecting DB.
$dbname = "filters";
mysql_select_db($dbname);
// 3. Build/Test SQL Query
$sql = ("select * from filter_bandpass where start_passband=" . $_POST['Lowfreq'] . " and stop_passband='" . $_POST['Highfreq'] . "'");
//echo $sql; //Comment/Uncomment to test sql query.
// 4. Retrieve info from MySQL.
$query = mysql_query($sql);
// 5. Display Query.
echo "<table border='1'>
<tr>
<th>Low Frequency</th>
<th>High Frequency</th>
</tr>";
while($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['Lowfreq'] . "</td>";
echo "<td>" . $row['Highfreq'] . "</td>";
echo "</tr>";
}
echo "</table>";
?>
Any help would be appreciated, I'm sure it's going to be some small stupid error i've over looked.
Thanks in advance :)
I'm guessing, based on your query, that you need to change this
mysql_select_db($dbname);
to
mysql_select_db($dbname, $connection);
and
while($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['Lowfreq'] . "</td>";
echo "<td>" . $row['Highfreq'] . "</td>";
echo "</tr>";
}
to
while($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['start_passband'] . "</td>";
echo "<td>" . $row['stop_passband'] . "</td>";
echo "</tr>";
}
Change line
mysql_select_db($dbname);
to
mysql_select_db($dbname, $connection);
Also before query check
$_POST['Lowfreq'] and $_POST['Highfreq']
If there is no value in these variables query will must return empty.
In your query there should not brackets for string.
$sql = ("select * from filter_bandpass where start_passband=" . $_POST['Lowfreq'] . " and stop_passband='" . $_POST['Highfreq'] . "'");
should be:
$sql = "select * from filter_bandpass where start_passband=" . $_POST['Lowfreq'] . " and stop_passband='" . $_POST['Highfreq'] . "'";

PHP - SQL select Data display

<html>
<head>
<meta http-equiv = "content-type" content = "text/html; charset = utf-8" />
<title>Using file functions PHP</title>
</head>
<body>
<h1>Web Development - Lab05</h1>
<?php
require_once("settings.php");
$dbconnect = #mysqli_connect($host, $user, $pswd, $dbnm);
if($dbconnect->connect_errno >0)
{
die('Unable to connecto to database [' . $db->connect_error . ']');
}
$queryResult = "SELECT car_id, make, model, price FROM cars";
echo "<table width='100%' border='1'>";
echo "<tr><th>ID</th><th>Make</th><th>Model</th><th>Price</th></tr>";
//initiate array
$displayrow= mysqli_fetch_array($queryResult);
//initiate while loop to iterate through table
while($displayrow)
{
echo "<tr>";
echo "<td>" . $row['ID'] . "</td>";
echo "<td>" . $row['Make'] . "</td>";
echo "<td>" . $row['Model'] . "</td>";
echo "<td>" . $row['Price'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($dbconnect);
?>
</body>
</html>
This is doing my head in, I cannot figure out why it will not display the actual data apart from the Table header. No matter what I used.
I have tried mysqli_fetch_array, mysqli_fetch_row, mysqli_fetch_assoc but nothing works.
Help and explanation why it was not displaying the data would be much appreciated :)
First: You aren't running a query, you are only putting the query text in a variable. You need to use mysqli_query.
Second: You should add mysqli_fetch_array to the loop.
For example:
while($displayrow = mysqli_fetch_array($queryResult))
{
}
Otherwise you are only getting the first row.
Third: Array keys are case sensitive. There is no $row['ID'], as Jeribo pointed out, it is $row['car_id'] as referenced in your query. $row['Make'] is not the same as $row['make'].
Please Precision to names of field in Query ( car_id,make,...)
while($displayrow= mysql_fetch_assoc($queryResult) )
{
echo "<tr>";
echo "<td>" . $displayrow['car_id'] . "</td>";
echo "<td>" . $displayrow['make'] . "</td>";
echo "<td>" . $displayrow['model'] . "</td>";
echo "<td>" . $displayrow['price'] . "</td>";
echo "</tr>";
}
If you want to query outside you still have to set it in the loop:
$result = $db->query($queryResult)
while($row = $result ->fetch_assoc()){
...
}
a Good Tutorial is shown here: http://codular.com/php-mysqli
$row needs to be initialized so why don't you try:
while($row = mysqli_fetch_array($queryResult))
{
....
}
You have to get the result set first and then try fetching array from result set
<?php
require_once("settings.php");
$dbconnect = #mysqli_connect($host, $user, $pswd, $dbnm);
if($dbconnect->connect_errno >0)
{
die('Unable to connecto to database [' . $db->connect_error . ']');
}
$query = "SELECT car_id, make, model, price FROM cars";
$resultSet=mysqli_query($dbconnect,$query)
echo "<table width='100%' border='1'>";
echo "<tr><th>ID</th><th>Make</th><th>Model</th><th>Price</th></tr>";
//initiate array
$displayrow= mysqli_fetch_array( $resultSet);
//initiate while loop to iterate through table
while($displayrow)
{
echo "<tr>";
echo "<td>" . $row['ID'] . "</td>";
echo "<td>" . $row['Make'] . "</td>";
echo "<td>" . $row['Model'] . "</td>";
echo "<td>" . $row['Price'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($dbconnect);
?>
http://www.w3schools.com/php/func_mysqli_fetch_array.asp

Fetch data from DB on same page

I am trying to let the user input an account number and below the business name will show.
Here is the code I am using which is called from the form:
<?php
$q = intval($_GET['q']);
$con = mysql_connect("localhost","cl49-xxx","xxx");
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"cl49-xx");
$sql="SELECT * FROM member WHERE personID = '".$q."'";
$result = mysqli_query($con,$sql);
echo "Business Name:<table>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['businesstype'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
When I run this code it just shows business name:.
I think you have issue near while loop. Instead of "while($row = mysqli_fetch_array($result)) ". Why do not you use this:
$row = mysqli_fetch_array($result);
$i= 0;
while( $i<count($row))
{
echo "<tr>";
echo "<td>" . $row['businesstype'] . "</td>";
echo "</tr>";
$i++;
}
echo "</table>";

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