I am trying to make the results from mysql results clickable links in php. I am new to php and please help.
<?php mysql_connect("localhost", "root", "pass") or die(mysql_error());
mysql_select_db("Branches") or die(mysql_error());
$data = mysql_query("SELECT * FROM items order by ID desc Limit 5") or die(mysql_error());
while ($info = mysql_fetch_array($data)) {
echo $info['title'];
echo " <br>";
echo $info['descr'];
echo "<br>";
echo "<br>";
} ?>
while ($info = mysql_fetch_array($data)) {
echo ''.$info['title'].'';
echo " <br>";
echo $info['descr'];
echo "<br>";
echo "<br>";
}
Then in somefile.php, use $_GET to capture the id and show the results
$id = $_GET['id'];
// pull info from db based on $id
$sql = mysql_query('SELECT * FROM items WHERE ID = "'.$id.'"');
.
.
.
.
just use anchor tag like this
while ($info = mysql_fetch_array($data)) {
echo "".$info['title'];. "";
}
echo '$info['descr']'
Related
I'm pretty new at PHP/MySQL, so please be patient with me.
I am trying to get a list of members in a table to show up on a page. Right now it's showing the first member about 10 times and not displaying anyone else's name. I DID have it working, but I don't know what happened. I just want it to display everyone's name once. Here is my code:
<?php $select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$row = mysql_fetch_array($select);
$rows = mysql_num_rows($select);
$teaching = $row[teaching];
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
$row2 = mysql_fetch_array($select2);
$student=$row2[student_name];
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
}
else {
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
$row3 = mysql_fetch_array($select3);
while($row2 = mysql_fetch_array($select2)) {
$house=$row3[house];
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>"; }
?>
I miss look on your code, since it is mess, but disregard the mysqli and mysql thing, you want to show how many student in the teacher's classes.
<?php $select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$row = mysql_fetch_array($select);
$rows = mysql_num_rows($select);
$teaching = $row[teaching]; <--- This only get first row of the course, if you want multiple course under same username, you need to loop it.
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
$row2 = mysql_fetch_array($select2);
$student=$row2[student_name]; <----- This only get the first row of the student name, if you want multiple student under a course, you need to loop it.
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
}
else {
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
$row3 = mysql_fetch_array($select3);
while($row2 = mysql_fetch_array($select2)) {
$house=$row3[house]; <----This only show the first row of $house under same student, so you need to loop it too.
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>"; }
?>
So what you really want to do is
<?php
$select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$rows = mysql_num_rows($select);
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
while( $row = mysqli_fetch_array( $select ) ) {
$teaching = $row[teaching];
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
} else {
while( $row2 = mysql_fetch_array($select2) ) {
$student=$row2[student_name];
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
while($row3 = mysql_fetch_array($select3)) {
$house=$row3[house];
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>";
}
} // END ELSE
}
} // END ELSE
?>
I'm trying to get data from a database if a link is clicked.
I used the example codes suggested from this example -Getting mysql field data when a link is clicked?
But it doesn't work when I click on a link nothing comes up.
main.php
<?php
include('conn.php');
$sql2 = "SELECT Title FROM addpromo";
$result2 = mysql_query($sql2);
echo "<div id=\"links\">\n";
echo "<ul>\n";
while ($row2 = mysql_fetch_assoc($result2)) {
echo "<li> <a href=\"fullproject.php?title=\""
. urlencode($row2['Title']) . "\">"
. htmlentities($row2['Title']) . "</a>\n</li>";
}
echo "</ul>";
echo "</div>";
?>
This is displaying correct.but when I click at a link nothing is showing up in fullproject.php, Just a blank page.
fullproject.php
<?php
// Connect to server.
include('conn.php');
$projectname = isset($_GET['Title']);
$sql1 = "SELECT Title FROM addpromo WHERE Title = '$projectname'";
$result1 = mysql_query($sql1);
while ($row1 = mysql_fetch_assoc($result1)) {
echo "Project Name: " . $row1['Title'] . "<br />";
echo "<br /> ";
}
?>
Can someone help me to fix this, or any other way to make this(to get data from a database if a link is clicked) possible?
Change to this
main.php
<?php
include('conn.php');
$sql2="SELECT Title FROM addpromo";
$result2=mysql_query($sql2);
echo '<div id="links">';
echo '<ul>';
while($row2 = mysql_fetch_assoc($result2)){
echo '<li>'.htmlentities($row2['Title']).'</li>';
}
echo '</ul>';
echo '</div>';
?>
fullproject.php
<?php
if(isset($_GET['title'])){
include('conn.php');
$projectname= $_GET['title'];
$sql1="SELECT Title FROM addpromo WHERE Title = '$projectname'";
$result1=mysql_query($sql1);
while($row1 = mysql_fetch_assoc($result1)) {
echo "Project Name: " . $row1['Title']. "<br />";
echo "<br /> ";
}
}
?>
This is storing a boolean value $projectname= isset($_GET['Title']);, whether or not the title is set. Instead use $projectname = $_GET['Title'];
isset returns a boolean value (true/false) and you want the actual value of the variable:
$projectname= $_GET['title'];
Furthermore, you have to pass only the title as the URL parameter, without enclosing it within quotes. So there is an error in this line:
echo "<li> <a href=\"fullproject.php?title=" . urlencode($row2['Title']) . "\">"
Note the lack of \" after title=
Can someone take a look at my code? I am trying to display a record from a link from the previous page. When I echo $id (below)... it displays the value of 'id' from the previous page. However the query returns all rows of the table. I had it working earlier. I have since changed something. Pleas point out what I over looking. Thank you, in advance.
<?php
mysql_connect('host', 'userid', 'password') or die (mysql_error());
mysql_select_db('my_database') or die (mysql_error());
//$result = mysql_query("SELECT * from table");
$id= mysql_real_escape_string($_GET["id"]);
echo $id;
$result = mysql_query("SELECT * FROM uatbeta1 WHERE id=".intval($_REQUEST['id']));
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result);
echo $result;
while($row = mysql_fetch_array($result)){
//Display the results from the current row and a line break
echo "<table border='1'>";
echo "<tr>";
echo "<td>".$row['id']."</td>";
echo "<td>".$row['company']."</td>";
echo "<td>".$row['constructioncontractor']."</td>";
echo "<td>".$row['weldingcontractor']."</td>";
echo "<td>".$row['ndtcontractor']."</td>";
echo "<td>".$row['weldid']."</td>";
echo "<td>".$row['projectname']."</td>";
echo "<td>".$row['examinationdate']."</td>";
echo "<td>".$row['shift']."</td>";
echo "</tr>";
Why not use $id? You have it nicely escaped, so use that variable:
$result = mysql_query("SELECT * FROM uatbeta1 WHERE id=".$id);
I'm new at php and JavaScript so I hope you can help me!
below is a php code! it take data from my db and display my photos and photo's title. what I want to do is if you click on one of this photos and than a new php expe: photo.php opens and you can see the same photo alone not with the other ones.
<?php
$link = mysql_connect("localhost", "root","");
mysql_select_db("test", $link);
$query = "select * from post order by id DESC;";
$results = mysql_query($query, $link) or die ("you comand can't be executed!".mysql_error());
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
echo $row['title']. "<img src=/1/" . $row['location']. " width=580px > " . "<br /><br /><br />";
}
}
?> '
<?php
$link = mysql_connect("localhost", "root","");
mysql_select_db("test", $link);
$query = "select * from post order by id DESC;";
$results = mysql_query($query, $link) or die ("you comand can't be executed!".mysql_error());
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
//send photo_id (id of photo from your table) from url NOTE: I've put link here
echo $row['title']. "<a href='index.php?photo_id=".$row['id']."'><img src=/1/" . $row['location']. " width=580px ></a> " . "<br /><br /><br />";
}
if(isset($_GET['photo_id'])) {
//here you get only one id of photo now retrieve the data from database and display the image
}
}
?>
You will get the photo by the help of the url value here photo_id.
do this way you are getting results from database and you are
displaying them with tile and image your aim is to click on
image or image title go to a new page and display image alone
the image title as link in first page
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
<a href='new page url?id=<?php echo $row["id"] ?>'>
echo $row['title'];
echo "</a>"
}
}
where $row['id'] is database unique id if have no it is always preferred
to create a database with unique id and in new page get the uid
using $id=$_GET['id'] and
do as below
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
if($id==$roe['id']{
echo $row['title']. "<img src=/1/" . $row['location']. " width=580px > ";
}
}
}
I have a large database of venues - and I would like to display this data in one page that would only change in some sort of an attribute to the id: (Ex: venues.php?id=1, which would get all the data from row #1.)
Edit: Okay, I updated the code and this is what it looks like now:
<?php
mysql_connect("localhost", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".(int)$id) ;
or die(mysql_error());
Print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$info['VENUE_NAME'] . "</td> ";
Print "<th>Address:</th> <td>".$info['ADDRESS'] . " </td></tr>";
}
Print "</table>";
?>
And upon going to venues.php?id=1 I get this error:
Parse error: syntax error, unexpected T_LOGICAL_OR in
/home/nightl7/public_html/demos/venues/venues.php on line 8
Do you mean something like:
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".(int)$id) ;
In order to "pass" the id into your url "venues.php?id=1"
You need to use a hybrid html/php form with method=get.
You can see an example html form here: w3schools html forms
This is what I would do:
print '<form name="input" action="venues.php" method="get">';
print 'Venue: <select name = "id">';
$con = mysql_connect("","","");
mysql_select_db($dataBase);
if (!$con){die('Could not connect: ' . mysql_error());}
else {
$opt = array();
$optVal = array();
$i = 0;
$sql = "Select * from venues";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
$opt[$i] = $row['VenueName'];
$optVal[$i] = $row['VenueID'];
print "<option value='$optVal[$i]'>$opt[$i]</option>";
$i++;
}
}
mysql_close($con);
print '</select><br />';
print '<input type="submit" value="Submit" />';
print '</form>'
This will give you a form that will give you a drop down list of all your venues and once a venue is selected will direct you to the venues.php page with the respective id.
at the top of your venues,php page just use
$id = $_GET['id'];
This assigns the id number to the variable $id and then you can use this "select"
$data = mysql_query("SELECT * FROM venues WHERE id = ".$id) ;
To collect your venue name from your database using the id supplied in the form.
Good Luck :)
<?php
mysql_connect("localhost", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".$id) or die(mysql_error());
Print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$info['VENUE_NAME'] . "</td> ";
Print "<th>Address:</th> <td>".$info['ADDRESS'] . " </td></tr>";
}
Print "</table>";
?>