jQuery & JSON replace all from string

[Mudsaf]

Hello, i'm wondering how to replace all data from content not just the amount $.each function gives.

i = 0;$.each(<JSON CODE>, function() {data = data.replace(<REPLACE CODE>,<REPLACE CODE>);i++;});

Example now the result from each is 2 and will replace only 2 items, but if there is more items on data how that works?

[justsomeguy]

The each method will loop through an array or the properties of an object. If it loops twice then either you have an array with 2 elements or an object with 2 properties. I'm not sure what your question is.

[Mudsaf]

Example my data would be like 500 characters long and might contain 10x strings i want to replace, without me knowing the amount how many strings there is & have to be replaced.

<script>var string = "THIS CAN BE REALLY LONG STRING :I: :I: asdas :I:"; //IDK HOW MANY :I: strings there would be//Replace them all</script>

[justsomeguy]

You can do a global replace with a regular expression in String.replace.

 

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace

[Mudsaf]

You can do a global replace with a regular expression in String.replace.

 

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace

data = data.replace(/<myjson>/g,<mystring>); //Didin't work.

[justsomeguy]

That's right, that's not the correct way to use String.replace. Look at the documentation again.

[Mudsaf]

That's right, that's not the correct way to use String.replace. Look at the documentation again.

var str = "Twas the night before Xmas...";var newstr = str.replace(/xmas/i, "Christmas");print(newstr); //Returns Twas the night before Christmas...

Isn't it similar to mine?

 

NEW

var data = data.replace(/<myjson>/g,<mystring>); //Didin't work.

[justsomeguy]

It's similar, yeah. Are you trying to replace the string "<myjson>" with something else?

[Mudsaf]

It's similar, yeah. Are you trying to replace the string "<myjson>" with something else?

 

Yes, json array for example

 

my_list.listIndex[0].item

[justsomeguy]

So you're trying to replace the entire string, not searching for a substring inside of it. If you're replacing the entire string then you don't need to use replace, you just overwrite the variable. But you're not really trying to replace the entire string, are you?

 

Show the actual code you're using instead of replacing the values with tokens like <myjson>. The values are the important part, there's no reason to remove those.

[Mudsaf]

Well the values are not actually important.

data = data.replace(smiley_list.smileyList[i].smileyCode,"<img class='" + smiley_list.smileyList[i].smileyCode + "' class='sbSmiley' style='vertical-align:middle;' src='script/shoutbox/smileys/" + smiley_list.smileyList[i].smileyURL + "'>");

So bascially i want to replace all codes where is smiley_list.smileyList.smileyCode with src smiley_list.smileyList.smileyCode

[justsomeguy]

 

 

Well the values are not actually important.

Why would you say that? The values are the important part. You have the general syntax right, but the values are what you are telling it to actually do. The code you posted will work, but it's only going to make one replacement. If you want to replace all then the first argument (the value) needs to be a regular expression with the g flag instead of a string. I'm assuming that the smileyCode parameter contains a string and not a regular expression object.

[Mudsaf]

Yea, how i can turn the string to expression?

[justsomeguy]

Just create a RegExp object:

 

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

 

Note the table of special characters, if those characters are in the string you might have to escape them. This is why the value is necessary though, when you write <myjson> I can't tell if you're using a regular expression or a string or what.

[Mudsaf]

Just create a RegExp object:

 

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

 

Note the table of special characters, if those characters are in the string you might have to escape them. This is why the value is necessary though, when you write <myjson> I can't tell if you're using a regular expression or a string or what.

	var smiley = new RegExp(smiley_list.smileyList[i].smileyCode, "g");	var data = data.replace(smiley,"<img class='" + smiley_list.smileyList[i].smileyCode + "' class='sbSmiley' style='vertical-align:middle;' src='script/shoutbox/smileys/" + smiley_list.smileyList[i].smileyURL + "'>");Uncaught TypeError: Cannot call method 'replace' of undefined 

Edited August 9, 2013 by Mudsaf

[justsomeguy]

Sounds like data is undefined, like the error says.

[Mudsaf]

Alright i totally messed the page, now basically everything = smileys (70%)

 

My JSON

{"smileyList": [{ "smileyCode":"[SMILEN]" , "smileyURL":"SM0-001.png"}, { "smileyCode":"[SMILEC]" , "smileyURL":"SM0-002.png"} ]}

My JavaScript (JSON stored to global variable named smiley_list

 

Javascript

	i = 0;	$.each(smiley_list.smileyList, function() {	smiley = new RegExp(smiley_list.smileyList[i].smileyCode, "g");	data = data.replace(smiley,"<img class='sbSmiley' style='vertical-align:middle;' src='script/shoutbox/smileys/" + smiley_list.smileyList[i].smileyURL + "'>");	i++;	});	$("#shoutbox_content").html(data);		}

Seems like its replacing every s with the image? dafuq

Edited August 9, 2013 by Mudsaf

[justsomeguy]

In a regular expression, square brackets mean a character class. The regular expression /[sMILEN]/ means to match any S, M, I, L, etc. You need to escape the brackets:

 

/[sMILEN]/

 

I think technically you only need to escape the opening bracket.

[Mudsaf]

Well how i can turn easily [sMILEN] to /[sMILEN]/ ?

[justsomeguy]

You can use String.replace to replace it in Javascript, or you can replace it in PHP before sending it to Javascript.

[Mudsaf]

smiley = smiley_list.smileyList[i].smileyCode;smiley = smiley.replace("]","]");Returns [SMILEN] 

[justsomeguy]

It looks like you need to use ].

[Mudsaf]

It looks like you need to use ].

 

Finally it works, thanks a thousand times :>