i am trying to get xml data from my website. with this code
$xml=simplexml_load_file('http://www.mywebsite.com/dataxml.php');
foreach ($xml->id as $id) {
echo "<tr>";
echo "<td>".$id->phone."</td>";
echo "<td>".$id->first_name." ".$id->last_name."</td>";
echo "<td>".$id->update."</td>";
echo "</tr>";
}
it work fine with below code in website file dataxml.php
header('Content-Type: text/xml');
include_once('customer.php');
$ThisCustomer = new Customer_Info;
echo $ThisCustomer->xmlCustomers();
but when i am trying to get with below code i get error.
session_start();
if(!isset($_SESSION['authorize'])) return;
// if i place this line before/after header result is same
header('Content-Type: text/xml');
include_once('customer.php');
$ThisCustomer = new Customer_Info;
echo $ThisCustomer->xmlCustomers();
Warning: simplexml_load_file(): http://www.mywebsite.com/dataxml.php:1: parser error : Document is empty in C:\wamp\www\xml_show.php on line 29
also note:
if i visit direct site with 2nd code like http://www.mywebsite.com/dataxml.php it does display xml data correctly.
Related
I am very new to both php and xml. What I am trying to do in
php is read in xml from a call to a url, and then parse the xml.
(I can get this to work in the example below when $urlip = 'localfile.xml'
but not when I put in a url. Ive checked the url by going to it with my browser,
and I can see the xml. I also did a show source, copied it and then pasted the
xml into the localfile and that works fine.
What am I doing wrong in trying to get the xml from the url?
Thank you
The error being returned is:
Error loading XML Start tag expected, ‘<' not found
Here is my code snip it:
$urlip="test.xml";# for debugging since I cannot read from the url yet! not sure why....
if (($xml = file_get_contents($urlip))===false) {
echo "error fetching XML\n";
} else {
libxml_use_internal_errors(true);
$data = simplexml_load_string($xml,null,LIBXML_NOCDATA);
if (!$data) {
echo "Error loading XML\n";
foreach(libxml_get_errors() as $error) {
echo "\t", $error->message;
}
} else {
foreach ($data as $item) {
$type = $item->TAB_TYPE;
$number=$item->ALT_ID;
$title = $item->SHORT_DESCR;
$searchlink = $item->ID;
$rsite=$item->CATEGORY;
echo "type $type, number $number, title $title, search link $searchlink, site $rsite\n";
}
}
}
Most likely situation from what it looks like:
Your function queries the remote URL and returns you an empty string, which passes the condition of your 'if' statement.
After that - you try to pass the empty string into XML, but it cannot, so it gives you an error.
Your steps to solve it:
configure php to open remote urls as comments to your question state - url_fopen
use another way to get content from the URL - cURL library works well
I am trying to pull area of a page with AJAX.
In JS I have on click I pass href to PHP;
in PHP(located in tools):
<?php defined('C5_EXECUTE') or die("Access Denied.");
$path = ($_POST['path']);
$page = Page::getByPath($path);
$a = new Area('Main');
$ret = $a->display($page);
echo json_encode($ret);
?>
If I make:
echo json_encode($page);
I receive the page so everything working, But when I try to receive an Area I get this error:
concrete\elements\block_area_header_view.php on line 5
In this File I found this
$c = Page::getCurrentPage();
$areaStyle = $c->getAreaCustomStyleRule($a);
So as I understand $c is null that why I have this error how can I fix this??
This line of code:
$ret = $a->display($page);
...does not do what you think it does. The "display" function does not return the content -- instead it outputs it to the browser. So your json_encode($ret) is just encoding and echo'ing an empty variable.
To capture the displayed content and put it into a variable, you can use php's output buffering feature, like so:
ob_start();
$a->display($page);
$ret = ob_end_clean();
$content = file_get_contents('file.php');
echo $content;
nothing displays, expect when displaying the page sourcecode in browser the display is this
<? foreach(glob("folder/*.php") as $class_filename) { require_once($class_filename); } ?>
so it wont execute the script when getting the content..
file.php contains this code
<? foreach(glob("folder/*.php") as $class_filename) {
require_once($class_filename);
}
?>
and if I do next
$content = foreach(glob("folder/*.php") as $class_filename) { require_once($class_filename); } ?>
it complains about unexpected foreach...
is there a way to read the folder/.php files content to single $variable and then echo/print all folder/.php files to page where it should be?
thanks for help already.
Is that what you want to do ?
$content = '';
foreach (glob('folder/*.php') as $class){$content .= file_get_contents($class);}
echo $content;
What you're trying won't execute the contents of the "file.php", jsut display the contents of them on screen.
If you want to execute file.php, use eval ($content)
To capture the output, use something like:
ob_start(); // Don't echo anything but buffer it up
$codeToRun=file_get_contents('file.php'); // Get the contents of file.php
eval ($codeToRun); // Run the contents of file.php
$content=ob_get_flush(); // Dump anything that should have been echoed to a variable and stop buffering
echo $content; //echo the stuff that should have been echoed above
This question already has answers here:
HTML into PHP Variable (HTML outside PHP code)
(7 answers)
Closed 4 years ago.
Hi i'd like to store a dinamically generated(with php) html code into a variable and be able to send it as a reply to an ajax request.
Let's say i randomly generate a table like:
<?php
$c=count($services);
?>
<table>
<?php
for($i=0; $i<$c; $i++){
echo "<tr>";
echo "<td>".$services_global[$i][service] ."</td>";
echo "<td>".$services_global[$i][amount]."</td>";
echo "<td>€ ".$services_global[$i][unit_price].",00</td>";
echo "<td>€ ".$services_global[$i][service_price].",00</td>";
echo "<td>".$services_global[$i][service_vat].",00%</td>";
echo "</tr>";
}
?>
</table>
I need to store all the generated html code(and the rest) and echo it as a json encoded variable like:
$error='none';
$result = array('teh_html' => $html, 'error' => $error);
$result_json = json_encode($result);
echo $result_json;
I could maybe generate an html file and then read it with:
ob_start();
//all my php generation code and stuff
file_put_contents('./tmp/invoice.html', ob_get_contents());
$html = file_get_contents('./tmp/invoice.html');
But it sounds just wrong and since i don't really need to generate the code but only send it to my main page as a reply to an ajax request it would be a waste of resources.
Any suggestions?
You don't have to store it in a file, you can just use the proper output buffering function
// turn output buffering on
ob_start();
// normal output
echo "<h1>hello world!</h1>";
// store buffer to variable and turn output buffering offer
$html = ob_get_clean();
// recall the buffered content
echo $html; //=> <h1>hello world!</h1>
More about ob_get_clean()
if the data is so much expensive to regenerate then I would suggest you to use memcached.
Otherwise I would go regenerate it every-time or cache it on the frontend.
for($i=0;$i<=5;$i++)
{
ob_start();
$store_var = $store_var.getdata($i); // put here your recursive function name
ob_get_clean();
}
function getdata($i)
{
?>
<h1>
<?php
echo $i;
?>
</h1>
<?php
ob_get_contents();
}
Hi I have been having problems with the google weather api having errors Warning: simplexml_load_string() [function.simplexml-load-string]: Entity: line 2: parser error ....
I tried to use the script of the main author(thinking it was my edited script) but still I am having this errors I tried 2
//komunitasweb.com/2009/09/showing-the-weather-with-php-and-google-weather-api/
and
//tips4php.net/2010/07/local-weather-with-php-and-google-weather/
The weird part is sometimes it fixes itself then goes back again to the error I have been using it for months now without any problem, this just happened yesterday. Also the demo page of the authors are working but I have the same exact code any help please.
this is my site http://j2sdesign.com/weather/widgetlive1.php
#Mike I added your code
<?
$xml = file_get_contents('http://www.google.com/ig/api?weather=jakarta'); if (! simplexml_load_string($xml)) { file_put_contents('malformed.xml', $xml); }
$xml = simplexml_load_file('http://www.google.com/ig/api?weather=jakarta');
$information = $xml->xpath("/xml_api_reply/weather/forecast_information");
$current = $xml->xpath("/xml_api_reply/weather/current_conditions");
$forecast_list = $xml->xpath("/xml_api_reply/weather/forecast_conditions");
?>
and made a list of the error but I can't seem to see the error cause it's been fixing itself then after sometime goes back again to the error
here is the content of the file
<?php include_once('simple_html_dom.php'); // create doctype $dom = new DOMDocument("1.0");
// display document in browser as plain text
// for readability purposes //header("Content-Type: text/plain");
// create root element
$xmlProducts = $dom->createElement("products");
$dom->appendChild($xmlProducts);
$pages = array( 'http://myshop.com/small_houses.html', 'http://myshop.com/medium_houses.html', 'http://myshop.com/large_houses.html' ) foreach($pages as $page) { $product = array(); $source = file_get_html($page); foreach($source->find('img') as $src) { if (strpos($src->src,"http://myshop.com") === false) { $product['image'] = "http://myshop.com/$src->src"; } } foreach($source->find('p[class*=imAlign_left]') as $description) { $product['description'] = $description->innertext; } foreach($source->find('span[class*=fc3]') as $title) { $product['title'] = $title->innertext; } //debug perposes! echo "Current Page: " . $page . "\n"; print_r($product); echo "\n\n\n"; //Clear seperator } ?>
When simplexml_load_string() fails you need to store the data you're trying to load somewhere for review. Examining the data is the first step to diagnose what it causing the error.
$xml = file_get_contents('http://example.com/file.xml');
if (!simplexml_load_string($xml)) {
file_put_contents('malformed.xml', $xml);
}