I have the next code:
<form method="POST">
<?php
echo "<select value>";
$xml = new SimpleXMLElement("http://news.google.com/news?ned=us&topic=h&output=rss",null,true);
foreach($xml->channel->item as $news)
{
echo "<option value='".$news->title."'>" . $news->title . "</option>";
}
echo "</select>";
?>
</form>
So i have a dropdown box with news titles, when i select a title i want to appear the description from the xml file and i dont know how to do it, if someone can help me, please
Please check this example and let me know if you want any more help.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$(".feeds").change(function(){
var val=$(this).val();
$(".common").hide();
$("."+val).show();
});
});
</script>
<style>
.common{display:none;}
</style>
<?php
$xml = new SimpleXMLElement("http://news.google.com/news?ned=us&topic=h&output=rss",null,true);
$select_html="<select class='feeds'>";
$d_html="";
foreach($xml->channel->item as $news){
$d_html.="<div class='".$news->guid." common'>".$news->description.'</div>';
$select_html.="<option value='".$news->guid."'>" . $news->title . "</option>";
}
$select_html.="</select>";
echo $select_html;
echo $d_html;
?>
Related
How to get the correct image instead of getting the first image using jquery, whats wrong with the code below, can anyone help me out. it is also alerting undefined in alert box
This is My php page which is recieving multiple images file from the database using php mysql
<html>
<head>
<title></title>
<!-- <script type="text/javascript" src="food_test.js"></script> -->
<script type="text/javascript" src="jquery/jquery-3.3.1.min.js">
</script>
</head>
<body>
<div id="showImages">
</div>
This pice of code below loops through the images from the database
<?php
for($i = 0; $i < $dbImg->count(); $i++){
$imgId = $dbImg->results()[$i]->gal_img_id;
$imgName = $dbImg->results()[$i]->gal_img_name;
echo "<a id='imgLink' href='image_gallery.php'>";
echo "<img id='galleryImage' width='100px' height='100px'
src='../images/pages/gallery/". $imgName . "'></img>";
echo "<input type='hidden' id='" . $imgId . "' name='gal_img_id'
value='" . $imgName . "'></input>";
echo $imgName;
echo "</a>";
}
?>
Here is my jquery code which is displaying only first image whenever i click any image.
$("a#imgLink img#galleryImage").each(function(){
$(this).click(function(){
$("#showImages").html($("input[name='gal_img_id']").val());
alert($("input[name='gal_img_id']").val());
});
});
</script>
</body>
</html>
It can be due to single quotes.
$("#showImages").html($("input[name=gal_img_id]").val());
alert($("input[name=gal_img_id]").val());
Try this. I had a error like this then i removed quotes from the input[name=example] and it worked.
I want to write Jquery code inside PHP tag. I wanted to do this so that I can perform AJAX after that.
I have tried to echo it, but it doesn't work.
<?php
require("conn.php");
$rs = mysql_query("select * from food order by LENGTH(price), price");
if($rs!=false && mysql_num_rows($rs)>0){
$counter ++;
while($row = mysql_fetch_array($rs)){
echo '
<script src="jquery.js">
<script>
$(document).ready(function(){
$("#'.$row["code"].'").click(function(){
echo "clicked";
});
});
</script>
';
}
mysql_free_result($rs);
}else{
echo mysql_error();
}
mysql_close();
?>
Here's a small rearrangement of your code.
<?php
require("conn.php");
$rs = mysql_query("select * from food order by LENGTH(price), price");
if($rs!=false && mysql_num_rows($rs)>0){
$counter ++;
while($row = mysql_fetch_array($rs)){
echo '<a class="row_food">'. $row["code"] .'</a>';
}
mysql_free_result($rs);
}else{
echo mysql_error();
}
mysql_close();
?>
<script src="jquery.js"></script>
<script>
$(document).ready(function(){
$('.row_food').click(function(){
alert("clicked");
return false;
});
});
</script>
You don't have to echo the javascript code every time in the loop! You only need to echo the content from your db inside the loop.
In the javascript/jquery, you could bind the click event to the element and do whatever you needed.
Hope it helps.
Just put </script> after jquery.js.
I currently have some php script that outputs results from a query. I would like to add at the end two buttons that will show/hide the final element, but am not sure how to do this.
The following is my php script:
while($result = mysqli_fetch_array($iname))
{
echo "<b>Event Name:</b> " .$result['EventName'];
echo "<br> ";
echo "<b>Location:</b> ".$result['Location'];
echo "<br>";
//this is where I would like to add my two buttons, that would show the "hidden" content when clicked
And here is what I have written in an HTML script, that I would like to reconcile into the PHP output:
<!DOCTYPE html>
<html>
<head>
<scriptsrc="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js></script>
<script>
$(document).ready(function(){
$("#hidden").hide();
$("#hide").click(function(){
$("#hidden").hide(500);
});
$("#show").click(function(){
$("#hidden").show(500);
});
});
</script>
</head>
<body>
<button id="show">Show</button>
<button id="hide">Hide</button>
<p id="hidden">
Some Random Text that will be shown when the buttons are clicked
</p>
</body>
</html>
Any suggestions as to how this should be done?
How about if you get the number of result rows with $num_rows = mysql_num_rows($result);
Then, put a counter in your loop.
$counter = 1;
$theClass="";
while($result = mysqli_fetch_array($iname))
{
if ($counter == mysql_num_rows($result);){
$theClass="showHide";
}
echo "<div class='$theClass'>";
echo "<b>Event Name:</b> " .$result['EventName'];
echo "<br> ";
echo "<b>Location:</b> ".$result['Location'];
echo "<br>";
echo "</div>
$counter++;
}
Then, apply your javascript to the div whose class="showHide"
<?php
require ("init.php");
?>
<head>
<script src="ajax.js"></script>
<script src="common.js"></script>
</head>
<body>
<?php
$query = "SELECT * FROM User1";
$result = mysqli_query($connection, $query);
echo '<form> ';
echo "Select a Users:";
echo '<select name="users" onchange="showUser(this.value)">';
while ($row=mysqli_fetch_assoc($result)){
echo $row=['Username'];
echo '<option value="'.$row=['Username'].'">'.$row=['Username'].'</option>';
}
echo '</select></form>';
?>
<div id="txtHint"><b>User info will be listed here.</b></div>
</body>
the problem i am having is that the drop down box only shows array multiple times and does not show usernames from my database however the connection to my database is working as when tryed debugging it got it to echo out just one username
IT should be like this:
while ($row=mysqli_fetch_assoc($result)){
//echo $row=['Username']; //Invalied here
echo '<option value="'.$row['Username'].'">'.$row['Username'].'</option>';
}
No need that =sign.
Correct:
echo $row=['Username'];// This is unnecessary.
echo '<option value="'.$row=['Username'].'">'.$row=['Username'].'</option>';// what is = doing here?
To:
echo '<option value="'.$row['Username'].'">'.$row['Username'].'</option>';
I just asked another question here: global variable and reduce database accesses in PHP+MySQL
I am using PHP+MySQL. The page accesses to the database and retrieve all the item data, and list them. I was planning to open a new page, but now I want to show a pop div using javascript instead. But I have no idea how to utilize the variables of PHP in the new div. Here is the code:
<html>
</head>
<script type="text/javascript">
function showDiv() {
document.getElementById('infoDiv').style.visibility='visible';
}
function closeDiv() {
document.getElementById('infoDiv').style.visibility='hidden';
}
</script>
</head>
<body>
<ul>
<?php foreach ($iteminfos as $iteminfo): ?>
<li><?php echo($iteminfo['c1']); ?></li>
<?php endforeach;?>
</ul>
<div id="infoDiv" style="visibility: hidden;">
<h1><?php echo($c1) ?></h1>
<p><?php echo($c2) ?></p>
<p>Return</p>
</div>
</body>
</html>
"iteminfos" is the results from database, each $iteminfo has two value $c1 and $c2. In "infoDiv", I want to show the details of the selected item. How to do that?
Thanks for the help!
A further question: if I want to use, for example, $c1 as text, $c2 as img scr, $c1 also as img alt; or $c2 as a href scr, how to do that?
Try this:
<?php foreach ($iteminfos as $iteminfo): ?>
<li>
<a href="javascript:showDiv(<?php echo(json_encode($iteminfo)) ?>)">
<?php echo($iteminfo['c1']); ?>
</a>
</li>
<?php endforeach;?>
Also, modify showDiv to take your row data:
function showDiv(row) {
document.getElementById('infoDiv').style.visibility='visible';
document.getElementById('infoDiv').innerHTML = row['c1'];
}
Basically, you have to consider that the javascript runs in the browser long after the PHP scripts execution ended. Therefore, you have to embed all the data your javascript might need into the website or fetch it at runtime (which would make things slower and more complicated in this case).
Do you want a single info area with multiple items listed on the page and when you click an item the info area is replaced with the new content??? or you want a new info area for each item??
I see something along the lines of the first approach, so I will tackle the latter.
<?php
//do some php magic and get your results
$sql = 'SELECT title, c1, c2 FROM items';
$res = mysql_query($sql);
$html = '';
while($row = mysql_fetch_assoc($res)) {
$html .= '<li><a class="toggleMe" href="#">' . $row['title'] . '</a><ul>';
$html .= '<li>' . $row['c1'] . '</li><li>' . $row['c2'] . '</li></ul>';
$html .= '</li>'; //i like it when line lengths match up with eachother
}
?>
<html>
</head>
<script type="text/javascript">
window.onload = function(){
var els = document.getElementsByClassName("toggleMe");
for(var i = 0, l = els.length; i < l; i++) {
els[i].onclick = function() {
if(this.style.display != 'none') {
this.style.display = 'block';
} else {
this.style.display = 'none';
}
}
}
}
</script>
</head>
<body>
<ul>
<?php
echo $html;
?>
</ul>
</body>
</html>