HeyAwesomePeople Posted August 6, 2015 Share Posted August 6, 2015 Hello! I know this seems like an easy task, using strval or casting to string, but the API I am using WON'T accept anything but a hard coded string "". I figure there has to be a way to get around this... What I have here is a method used to convert an array in to a YAML/String format for me to upload onto a server using JSONAPI, found here. I am using Spyc, as that is the only thing I have found that works for this, and it does pretty well. function pushArrayToServer($array, $api) { $yaml = Spyc::YAMLDump($array,4,60); $value = 0; while (strpos($yaml, "$value:") !== false) { $yaml = str_replace("$value:", "-", $yaml); $value = $value + 1; } $yaml = str_replace("---", "", $yaml); // Final YAML from Array var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", $yaml))); } Now the problem is that last $api->call I do. This call accepts the method type as a string(argument 1), and an array for argument two. For the method "files.write", the api requires an array being (file location, string). The string is what will be replacing the content of the file. But the only way I can get that line to work is if I do this: var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", "a string here"))); That works 100% fine, no errors. This is the dump I get when I run that. But as soon as I do one of these: var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", $yaml))); $yaml2 = $yaml . ""; var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", $yaml2))); var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", strval($yaml)))); var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", "$yaml2"))); var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", (string)$yaml))); These do not work. The dumps all return null. I haven't found a way yet to not do "". It almost worked once when I attempted some things, but I found out I was just converting a boolean into a string which caused an error. The server I am working with is a minecraft server with JSONAPi installed. It works great, except for this error. So I am assuming the string type has to be just like Java's, or pretty plain? I have no clue this is the first real issue I've had with this plugin. Thanks in advance, HeyAwesomePeople Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/ Share on other sites More sharing options...
requinix Posted August 6, 2015 Share Posted August 6, 2015 You are definitely passing a string, so the problem must lie within the value of $yaml itself. Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518133 Share on other sites More sharing options...
HeyAwesomePeople Posted August 6, 2015 Author Share Posted August 6, 2015 You are definitely passing a string, so the problem must lie within the value of $yaml itself. Maybe, but what do I do then? Printing $yaml into a text file results in what I want. $myfile = fopen("newfile.txt", "w") or die("Unable to open file!"); fwrite($myfile, $yaml); fclose($myfile); Result: http://hastebin.com/ucurevadup.sm Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518166 Share on other sites More sharing options...
requinix Posted August 7, 2015 Share Posted August 7, 2015 And now what happens if you put that all into a literal string and call the API with it directly? You should get the same error. The next question is what did you put in the "a string here" that made the API call work, and how does it differ with $yaml? Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518199 Share on other sites More sharing options...
HeyAwesomePeople Posted August 8, 2015 Author Share Posted August 8, 2015 And now what happens if you put that all into a literal string and call the API with it directly? You should get the same error. The next question is what did you put in the "a string here" that made the API call work, and how does it differ with $yaml? THERE'S PARENTHESIS. Ooohhh..... Okay so I need those parenthesis in the YAML to make it work right, so how would I avoid creating an issue with those? Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518331 Share on other sites More sharing options...
HeyAwesomePeople Posted August 8, 2015 Author Share Posted August 8, 2015 THERE'S PARENTHESIS. Ooohhh..... Okay so I need those parenthesis in the YAML to make it work right, so how would I avoid creating an issue with those? Quotations * Not parenthesis... Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518332 Share on other sites More sharing options...
requinix Posted August 9, 2015 Share Posted August 9, 2015 What quotes? Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518347 Share on other sites More sharing options...
HeyAwesomePeople Posted August 10, 2015 Author Share Posted August 10, 2015 (edited) What quotes? The script fails because of the "" and '' in the yaml file... see here At least this is what I am assuming, because I can paste any part of the file but I have to escape the quotations. This works but then I have backslashes in my yaml file... Okay scratch what I just said here ^ If I use a literal string, like so, var_dump($api->call("files.write", array("plugins/GroupManager/worlds/world/groups.yml", " groups: Owner: default: false permissions: \"\" - \'*\' inheritance: \"\" - Coowner info: prefix: \'&4[Owner]\' build: true suffix: \"\" HeadAdmin: default: false permissions: [ ] inheritance: \"\" - g:server_head_admin - admin info: prefix: \'&c[Head Admin]\' build: true suffix: \"\" Admin: default: false permissions: [ ] inheritance: \"\" - g:bukkit_admin - g:towny_admin - g:groupmanager_admin - g:server_admin - Architect info: prefix: \'&c[Head Admin]\' build: true suffix: \"\" Architect: default: false permissions: [ ] inheritance: \"\" - g:server_architect - HeadModerator info: prefix: \'&a[Architect]\' build: true suffix: \"\" HeadModerator: default: false permissions: [ ] inheritance: \"\" - g:server_head_moderator - Moderator info: prefix: \'&5[Head Moderator]\' build: true suffix: \"\" Moderator: default: false permissions: [ ] inheritance: \"\" - g:bukkit_moderator - g:towny_moderator - g:server_moderator - Helper info: prefix: \'&d[Moderator]\' build: true suffix: \"\" Helper: default: false permissions: [ ] "))); This works fine(Other than the fact it adds backslashes to the yaml). The character count of the string is 1622 characters(this includes spaces). But if I add ANY character, just one more space or letter or number, the code fails and outputs this in its var_dump: null At 1622 characters I get a success message. array (size=1) 0 => array (size=4) 'result' => string 'success' (length=7) 'success' => boolean true 'source' => string 'files.write' (length=11) 'is_success' => boolean true So at first I think there's a limit of 1622 characters, but that's a very specific number. So I put into the literal string 3000 random characters and it worked.... I'm lost here. The difference between working and not working is 1 character...? I did another test too. I cut the yaml file I am editing to around 1550 characters and my script works perfectly fine all the way through. I figured this would be the case, but this means the quotes don't have anything to do with it. So I'm realllllyyyy confused. In the YAML format, I can only hit 1622 characters. Is there a php limit somewhere I don't know about? Edited August 10, 2015 by HeyAwesomePeople Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518383 Share on other sites More sharing options...
requinix Posted August 10, 2015 Share Posted August 10, 2015 The backslash problem is because you're putting backslashes in places that don't need it. Since the backslashes aren't necessarily there, PHP keeps them in place thinking that you wanted literal backslashes. Don't use them for single quotes in a double-quoted string, or vice versa. Whatever you're experiencing is not due to PHP. I'm more troubled by the "3000 random characters and it worked" statement, which must mean something besides other than what those words literally mean. This is probably the time when you should be looking more closely at the API itself. If the code returns null then look to see under what conditions it will return null. Trace it back up to the actual API call (or whatever) and find out exactly what's happening. Because poking and prodding with different strings here and there isn't making much progress. Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518394 Share on other sites More sharing options...
HeyAwesomePeople Posted August 10, 2015 Author Share Posted August 10, 2015 The backslash problem is because you're putting backslashes in places that don't need it. Since the backslashes aren't necessarily there, PHP keeps them in place thinking that you wanted literal backslashes. Don't use them for single quotes in a double-quoted string, or vice versa. Whatever you're experiencing is not due to PHP. I'm more troubled by the "3000 random characters and it worked" statement, which must mean something besides other than what those words literally mean. This is probably the time when you should be looking more closely at the API itself. If the code returns null then look to see under what conditions it will return null. Trace it back up to the actual API call (or whatever) and find out exactly what's happening. Because poking and prodding with different strings here and there isn't making much progress. If I don't escape the "" and '' I get php errors. Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518426 Share on other sites More sharing options...
requinix Posted August 10, 2015 Share Posted August 10, 2015 If I don't escape the "" and '' I get php errors.One of them. You escape one of them. The one that you are using for the string delimiters. Not both of them. Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518428 Share on other sites More sharing options...
HeyAwesomePeople Posted August 11, 2015 Author Share Posted August 11, 2015 One of them. You escape one of them. The one that you are using for the string delimiters. Not both of them. Oh yeah... Okay so escaping just the "" doesn't result in an error, I just get the null message again. And the API is pretty simple. https://github.com/alecgorge/jsonapi http://hastebin.com/jozigaruru.php The API provides a way to communicate between a Minecraft server, Java, and the user. This can be through web app or iphone/android app or whatever. I don't think it's the API's fault, as Adminium(IOS app) also has a file manager system using the JSONAPI, and their files work just fine(including the types of one I am messing with). I looked through the Java anyways and found the method I am calling through the website: http://hastebin.com/avodositud.coffee Now this method only returns a success message or an error message, not nothing. It also state here that it should only return true or an error message... I'm also not the best at PHP(still fairly new to it) so the JSONAPI php file is hard to read for me, but I don't see anywhere I could get an error.. I cannot figure this out still. Thanks for helping me so far. What else should I be trying/looking at? Quote Link to comment https://forums.phpfreaks.com/topic/297659-converting-variable-into-a-string/#findComment-1518446 Share on other sites More sharing options...
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