I am running into a problem where the below is not working as I think it should.
// this will not work and $usernames appears empty
$usernames = "\"user1\", \"user2\"";
$query->notContainedIn("username", [$usernames]);
// this does work
$user1 = "user1";
$user2 = "user2";
$query->notContainedIn("username", ["$user1", "$user2"]);
Thanks
According to https://parse.com/docs/php_guide#queries you need to provide an array and $usernames is a string. By using the short-array notation, i.e. the square brackets, you've created an array with precisely one element, namely the string. What you need is an array with two elements corresponding to the two values of interest.
If for some reason your data is only available in this type of formatted string, then you would need to convert it to an array. There are at least two ways you could do this:
<?php
$usernames = explode(",",str_replace("\"","","\"user1\",\"user2\""));
?>
This is less complicated than it might appear. The code replaces the quotes encompassing each value with empty strings. Then it explodes the string in two by indicating that the comma is a delimitor. Although I've never worked with Parse, I would expect that you could replace the second parameter in the query with $usernames now, as follows:
<?php
$query->notContainedIn("username", $usernames);
Don't use the square brackets or you'll be creating a multidimensional array whose first element contains the array $usernames.
Alternative way of converting said string to array:
<?php
$string = "\"user1\",\"user2\"";
$tok = strtok($string, "\",");
$usernames = null;
while ($tok !== false) {
$usernames[] = "$tok";
$tok = strtok("\",");
}
$usernames now contains two values just like in the previous solution I proposed. While this second solution is more involved, note that it uses only one function instead of two, so it might produce faster results. You could try both solutions, benchmark them, and then pick the one you find to have superior performance.
Hope this helps.
Related
$myArray = array();
for($i=0;i<2;$i++){
..
.. //get the content logic here
assign it to array
$myArray["item"]["content"] = $item_content;
}
echo json_encode($myArray);
The above code produced this result:
Which has an error because I didn't merge them.
I tried to merge like this:
$finalJson = array_merge($finalJson, $myArray);
but the output of echo $finalJson is one object instead of 3.
Update:
Your real problem is down to your use of array_merge on an associative array. The behaviour of array_merge is to reassign duplicate keys if they are associative (cf the docs):
If the input arrays have the same string keys, then the later value for that key will overwrite the previous one. If, however, the arrays contain numeric keys, the later value will not overwrite the original value, but will be appended.
And because $myArray is clearly using strings as keys, the values in $finalJson are being reassigned. What you need to do is either create unique keys on each iteration, or simply push the values of $myArray onto a numerically indexed $finalJson array (like I showed in my original answer below)
The problem is simply this:
$myArray["item"]["content"] = $item_content;
it's inside the loop, each time reassigning the value of $myArray["item"]["content"], and not adding it to the array. I suspect that what you wanted to do is add this at the bottom of your loop (for each new value of $myArray):
$finalJson[] = $myArray;
Then, on the next iteration, $myArray will be reassigned, and its new value will be appended to the $finalJson variable.
i have a tricky problem.
What i do.
I generate from the System Tables of Databases (DEV ,Test Prod setup) information for Tables, Vies trigger … with PHP and compare the results ti see teh differences with JavaScript.
Also I have a Documentation DB for additional business information which was installed only once on TEST DB.
Therfore I need to connect all four environments to get the data.
I use
if ($flag === 'i')
for information DB and
elseif ($flag === 's')
for system db‘s.
Result is $row_array and $info_array which must be combined and sendet back to Javascript.
$json_response = array_merge($rinfo, $rsql );
echo json_encode($json_response, JSON_UNESCAPED_UNICODE | JSON_UNESCAPED_SLASHES | JSON_NUMERIC_CHECK);
I try this merge in a different positions in the program.
First time in
elseif ($flag === 'i') {
result json:
[
{
"0": "ACT",……….
} ][ ]
second time after } //ifelse return also 2 arrays
[{"0":"ACT","1":"Tabelle Akten","2":"hh","3":null,"4":null,"5":"UCC","6":"Y","7":"Reload Data in Test","8":"y","9":"delete all older tha","10":"n","11":" ","12":"y","13":" ","14":"o","15":"y","16":"o","17":"y","18":"y","19":"Diese tabelle speichert die Acten Verweise","20":"Gert Dorn","21":1570254359,"TDESCRIPTION":"n","TTYPE":" ","TREC_ESTIM":"y","TREC_GROWTH":" ","TDOMAIN":"o","TREL_TYPE":"y","TREL_RULES":"o","THOUSEKEEPING":"y","THOUSE_RULES":"y","TCID":"Diese tabelle speichert die Acten Verweise","TCID_RULES":"Gert Dorn","TUSE_UCC":1570254359,"TUSE_DWH":"","TUSE_ODS":"","TUSE_CWF":"","TUSE_IWF":"","TUSE_OWF":"","TUSE_DEP_MANAGER":"","TENTITY_DESCRIPTION":"","TOWNER":""
,"TTIMESTAMP":""**}][{**"0":"ACT","1":"DB2INST1"
,"2":"USERSPACE1","3":null,"4":"2018-11-21 16:43:20.066567","5":"2018-12-07 10:12:10.255759","6":null,"7":"2020-03-26","8":"2018-11-21 16:43:20.343258","9":3,"NAME":"ACT","CREATOR":"DB2INST1","TBSPACE":"USERSPACE1","REMARKS":"","CTIME":"2018-11-21 16:43:20.066567","STATS_TIME":"2018-12-07 10:12:10.255759","STATISTICS_PROFILE":"","LASTUSED":"2020-03-26","ALTER_TIME":"2018-11-21 16:43:20.343258","COLCOUNT":3}]
The program code and result you can download at
http://dmdg.io/dmdg.zip
Hope you can help Kind regards gert
Did you consider using array_push?
array_push is always preferred than myArray[] = $value
Question is related with example from here http://lv1.php.net/array_merge
$beginning = 'foo';
$end = array(1 => 'bar');
$result = array_merge((array)$beginning, (array)$end);
Usually I use such code $result2 = array_merge( array($beginning), $end );
$end is already an array. Why need (array)$end....
Tested and see the same result.
So question. Is array_merge( array($beginning), $end ) correct code?
Seems now understood why it is reasonable to use (array)
For example $var2 = array('test2');
print_r( array($var2) );
would be multidimensional array
but
print_r( (array)$var2 );
would be the same array as initial.
There is a slight difference between array($foo) and (array)$foo, but it won't affect the output.
While array($foo) will try to build an array out of $foo, obviously returning an array, (array)$foo will try to look at$foo like it is an array, hence returning an array. Both have the exact same result if your variable is a good candidate for an array, but (array)$foo may have a stronger semantic aspect since it exposes your intention of using the variable as an array, rather than building an array out of it.
array_merge only accepts parameters of type array (Since PHP 5.0)
Convert all parameters use typecasting, therefore
Add (array) before the variable, it's means convert the data type into array, case it is not array.
Note:
If you can ensure all of the variables which used in array_merge ARE array. You can direct access it, instead of adding the (array).
Yes, it's correct code. If you are sure that the parameter is already an array you don't need the type casting.
I'm trying to implode variables, but it's not working correctly:
$models = array("$model0, $model1");
$modelfinal = implode("," , $models);
$modelfinal only returns , ,
I'm guessing I'm way off...anybody?
The following statement creates an array with exactly one string in it, which is comprised of the values of two (apparently) undefined variables separated with a comma:
$models = array("$model0, $model1");
The end result is the same as if you had done this:
$models = array(", ");
Now you're imploding it using a comma as the separator, which doesn't do anything since there's only one element in the array (a string with a comma and a space).
Assuming $model0 and $model1 are defined (which is a problem you'll need to look into first), you can get your desired result either by:
directly using $modelfinal = "$model0, $model1",
or by using $models = array($model0, $model1); followed by the implode.
here is your problem "$model0, $model1" change it to this code
$models = array($model0,$model1);
I currently have var: $_REQUEST['fb_friend_uid'] which gives me the following output:
Array{"returned_val":["47483647","47483647","47483647","665414807","263901486","665414807","665414807","665414807"]}
Im looking to save the data here into a new array, containing only the numbers in a format of; 47483647, 47483647, etc
The objective is to use it in a sql query like so:
SELECT * FROM vote WHERE
vote_fb_uid IN ($myNumbers)
Saving it into a new array I figured could be done like so:
foreach ($_REQUEST['fb_friend_uid'] as $uid) {
$uids[] = $uid['id'];
}
$ids = join(',', $uids);
However my issue remains, how to "clean" the first var to contain numbers only.
Suggestions?
I can't give you an exact solution, because I'm not sure if the value returned by $_REQUEST['fb_friend_uid'] is a PHP array printed using json_encode(), or the value is actually a json string.
In either case, where is an example which makes use of both circumstances, so use whichever one makes sense in your scenario:
If PHP Array:
Assumes PHP Array has a format similar to:
array('returned_val' => array('47483647', '47483647', '47483647', '665414807', '263901486', '665414807', '665414807', '665414807'));
<?php
$original_arr = $_REQUEST['fb_friend_uid']['returned_val'];
If JSON String:
Assumes the JSON String has a format similar to:
{"returned_val":["47483647","47483647","47483647","665414807","263901486","665414807","665414807","665414807"]}
<?php
$json_arr = json_decode($_REQUEST['fb_friend_uid'], True);
$original_arr = $json_arr['returned_val'];
Then, use this code:
<?php
// Extract only whole number values, omit anything which is not a 0-9 character.
$filtered_arr = array_filter($original_arr, 'ctype_digit');
// Escape values to remove possibility of SQL injection.
$filtered_arr = array_map('mysql_real_escape_string', $filtered_arr);
// Convert the array to a string
$string_arr = "'" . implode("','", $filtered_arr) . "'";
// Perform SQL Query
mysql_query("SELECT * FROM vote WHERE vote_fb_uid IN ($string_arr)");
Just filter the array using is_numeric:
$uids = array_filter($_REQUEST['fb_friend_uid'], 'is_numeric');
To filter for numbers you can use is_numeric( mixed $var ).
But if you need more control (only integers of certain type, length) you can either use REGEX or is_numeric and some ifs.
This looks like a json string, so use http://php.net/json_decode
Maybe you need to remove Array at the beginning (but I don't know if Array is actual in the variable), use http://php.net/substr
$jsonString = substr($_REQUEST['fb_friend_uid'], 5);
$fb_friend_uid = json_decode($jsonString);
$ids = join(',', $fb_friend_uid['returned_val']);
I'm not 100% but this ($settings) would be called an array in php:
$setting;
$setting['host'] = "localhost";
$setting['name'] = "hello";
but what's the name for this that's different to the above:
$settings = array("localhost", "hello");
Also from the first example how can i remove the element called name?
(please also correct my terminology if I have made a mistake)
I'm not 100% but this ($settings)
would be called an array in php:
You should be 100% sure, they are :)
but what's the name for this that's
different to the above:
This:
$setting['host'] = "localhost";
$setting['name'] = "hello";
And this are different ways of declaring a php array.
$settings = array("localhost", "hello");
In fact this is how later should be to match the first one with keys:
$settings = array("host" => "localhost", "name" => "hello");
Also from the first example how can i
remove the element called name?
You can remove with unset:
unset($setting['name']);
Note that when declaring PHP array, do:
$setting = array();
Rather than:
$setting;
Note also that you can append info to arrays at the end by suffixing them with [], for example to add third element to the array, you could simply do:
$setting[] = 'Third Item';
More Information:
http://php.net/manual/en/language.types.array.php
As sAc said, they are both array. The correct way to declare an array is $settings = array(); (as opposed to just $settings; in your first line.)
The main difference between the first and second way is that the first allows you to use $settings['host'] and $settings['name'], whereas the latter can only be used with numeric indices ($settings[0] and $settings[1]). If you want to use the first way, you can also declare your array like this: $settings = array('host'=>'localhost', 'name'=>'hello');
More reading on PHP arrays
Well this is indeed an array. You have different types of array's in php. The first example you mention is called an Associative Array. Simply an array with a string as a key.
An associative array can be declared in two ways:
1) (the way you declared it):
$sample = array();
$sample["name"] = "test";
2)
$sample = array("name" => "localhost");
Furthermore the first example can also be used to add existing items to an array. For example:
$sample["name"][] = "some_name";
$sample["name"][] = "some_other_name";
When you execute the above code with print_r($sample) you get something like:
Array ( [name] => Array ( [0] => some_name [1] => some_other_name ) )
Which is very usefull when adding multiple strings to an existing array.
Removing a value from an array is very simple,
Like mentioned above, use the unset function.
unset($sample["name"])
to unset the whole name value and values connected to it
Or when you only want to unset a specific item within $sample["name"] :
unset($sample["name"][0]);
or, ofcourse any item you'd like.
So basicly.. the difference between your first example and the latter is that the first is an associative array, and the second is not.
For further reference on arrays, visit the PHP manual on arrays