I want to get the content of this 1st and 2nd webpage https://www.goodreads.com/search?utf8=%E2%9C%93&q=%D8%B3%D9%8A%D8%A7%D8%B3%D8%A9&search_type=books
and then store it in my database and then make the list is searchable.so after I googled I found that I can do it like this
$url = "https://www.goodreads.com/search?utf8=%E2%9C%93&q=%D8%B3%D9%8A%D8%A7%D8%B3%D8%A9&search_type=books";
function get_links($url){
$input = file_get_contents($url);
echo $input;
}
get_links($url);
my problem is how can I get the 2nd page content also and how can I store these books in my database to the list searchable
The answer is not that easy...
Options
Getting the pages (Not recommended)
To get a later page you can send the "page argument" in your request:
e.g.:
https://www.goodreads.com/search?page=2&q=%D8%B3%D9%8A%D8%A7%D8%B3%D8%A9&search_type=books&tab=books&utf8=%E2%9C%93
But to get the Elements into a nice structure you need to parse the HTML you get which is realy hard.
Use the API (recommended)
At https://www.goodreads.com/api/index you can find the documentation for goodreads API which returns example as response and is easily parseable.
Parse XML in PHP
If you use the API you can parse the XML-response with SimpleXML.
See example 1 & 2: http://php.net/manual/en/simplexml.examples-basic.php
Saving to database
If you are a beginner you might read some tutorials about how to use mysql with php and PDO. But you may also have a look at RedBeanPHP which is an really easy to use ORM for databases.
For getting 2nd page you will have to append on the url ?page=2
In case that you need to get other pages, you can use a for loop if you don't know how to use for loop, please google php for loop
For the database, google php mysql insert
I have to list the table entries from mysql in my iPhone app when a button is pressed. I am able to do it properly but it is just not the way I want it. I am using a PHP script to send Xcode what has to be printed.
First, I tried doing it using HTML table, but I didn't like the way it was printed with cells. Next I tried printing plain text by giving spaces(between columns) and \n for every new row. I used NSURL and loaded the webView to the iPhone. Looks good on browser but the same is not preserved when the iPhone tries to open it.
Is there a good way to do this? So I can just list the table entries without having to go through the traditional HTML table or any other idea is welcome.
Also, please try to be easily understood, as I am new to Obj-C, and PHP as well.
Thanks!!
Any thoughts on how I can do this in a UITableView..?? Do I have to return a string with component separation characters and fill in the tableView?
Output the results encoded in JSON. Send an a(sync) request to the server using NSURLConnection or using a third-party library such as AFNetworking. Parse the JSON using NSJSONSerialization, turning the results into an array/dictionary depending on the contents. Then parse the results into the UITableViewCell. It may be easier to subclass the cell so that you include the data that you'd like to use.
To encode the results from the database into JSON, you can use the method json_encode().
Hi friends i am working on JQ-GRID. I want to show Image in specific column, But i don't know how to attach image in JQ-GRID. Can anybody help me or please send me some links, thanks
jqgrid is a feature monster. I tell this everybody who asks about it.
When jqgrid loads, a function is called which actually gets the data you want to display.
This is normally an ajax call to your php. As a result set of this function, you can just use xml or json.
I prefer json, so I build my result array and do a echo json_encode($myarray)
jQuery("#your_grid_id").jqGrid({ url : '/ajax/getjqgriddata.php'})
Now displaying pictures, there are different ways you can do that. You can either generate a <img src="wherever/mypicture1.png"></img>-link and hand it over in your result, or encode your picture binary data with base64 and deliver it with your result.
A more addvanced way is to use an so called formatter and just returning a id for the image.
This depends on you, but I would suggest to get confident with jqgrid, experiment with returning -links to get a feeling how jqgrid works.
There is plenty of good documentation at:
http://trirand.com/blog/jqgrid/jqgrid.html
Just take a look at it.
I installed the JSON API Plug-In for my wordpress site. I want to display on an external site how many posts I have in a certain category.
For example, I have a URL like this: http://mywordpress.com/?json=get_category_posts&slug=press%20releases
Along the output, I almost immediately see a field "post_count" and a number right after it. I want to encapsulate this number into a string so I can output that number onto a table. But in order for me to be able to do that I need to have the entire output in a variable.
But I can't have something like
$json-output = http://mywordpress.com/?json=get_category_posts&slug=press%20releases;
echo $json-output;
That's just gonna output the URL, not the result of the JSON query.
I hope the question is making sense. Any guidance would be appreciated.
You will need to use CURL to get the json result from the URL, then parse the json with PHP, then echo the item you would like displayed.
The parts you seem to be missing are the CURL and JSON parsing, but you are on the right track.
This might help, a pretty simple example:
http://www.katcode.com/http-requests-using-curl-and-decoding-json-responses/
I just started doing jQuery last week, and so far I already made some basic systems with ajax, like basic jQuery CRUD and simple chat system without referencing on other's work for I decided to test myself on how far I can do systems alone in jQuery(without JSON and XML yet).
But when I decided to look at other's work (hoping to get/learn good practices and codes out there) many or almost every program that deals with ajax have some JSON in it. So I decided to study and read JSON specially this one, but I guess because it's my first time dealing with it, I'm having a problem sinking it into my brain. Yeah I know it is a "lightweight way of describing hierarchical data", I also know how to make JSON like mixing a literal array and object in JS, and how to dsplay it in js.
But my question is, what's the difference and what's the advantage than not using it?
When I can still get and store data on the server using ajax and database without JSON.
By the way I haven't focus on XML yet because based from my research it's better to use JSON in AJAX.
Can you give me some actual scenario dealing with
s1. ajax php mysql (this with what disadvantages?)
and
s2. ajax php mysql json (this with what advantages?)
I mean, my focus is to send and get data, and I already can do it with s1.
Sorry if you find my question stupid. Tia. :)
Why use JSON? The answer is portability and structure.
JSON is portable because parsers and writers are available for many, many languages. This means that JSON that a PHP script generates can be very easily understood by a JavaScript script. It is the best way to transmit complex structures like arrays and objects, and have it still be compatible with multiple languages.
JSON provides structure because the data you transmit with it can have consistent formatting. This is instead of transmitting back plain-text (i.e. unformatted) data, like comma-separated or delimited data.
Data that is merely delimited (for example, "BookName1,BookName2,BookName3") is more difficult for humans to understand, debug, and work with. If you wanted to debug a response between your server and your browser and the data was delimited (like my example above), you might have a hard time understanding it. Also, if you want to add different data types, provide separate records, etc., then your custom data format becomes more complicated. Eventually, you might end up reinventing JSON.
As a side note, JSON is indeed better than XML. It is much more efficient space-wise. There are no tag names to take up space. Structure is created via nested braces, instead of verbose tags.
Resources
Here is an interesting article on the differences and pros/cons of XML and JSON: http://www.json.org/xml.html
Examples
Per your request, here is an example of encoding JSON with PHP. This is ripped from the docs:
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo json_encode($arr);
Output:
{"a":1,"b":2,"c":3,"d":4,"e":5}
Contrast this to something like this, without JSON:
a,1
b,2
c,3
d,4
e,5
To parse that, you'd have to iterate through each line, split the values yourself, and then create the array. This isn't that difficult, but imagine you have a nested object:
$arr = array ('a'=> array(1,2,3),'b'=> array('a' => 1, 'b' => 2),'c'=>3,'d'=> array(1,2,3,4,5) ,'e'=>5); // etc.
With JSON, it's no different to encode it. Just use json_encode. But, encoding this manually, and then decoding it manually would be significantly more work.
Programming in any sort of programming language, you have several different types of data at your disposal, including the very useful array type.
Interchanging data between Javascript and any server side language can only happen through strings. I.e. you can send and return any text, but there's no way to send a native array or number type.
JSON is an elegant way to express array and other types using only a string. This way you can pass arbitrary data back and forth between different environments and are not limited to pure text. XML solves the same kind of problem, but is often overkill for simple AJAX requests.