modify json array recursively

Question

I have a json array like this:

$json = '[["Nome","contains","av"],"and",[["IdAreoporto","=",1],"and",["!",["IdAreoporto","=",3]]]]';

which basically is an sql. I need to obtain this:

$json = '((Nome like "%av%") and ((IdAreoporto = 1) and (not(IdAreoporto = 3))))';

so basically:

change parenthesis
replace commas with spaces
remove quotes except if array length=3 and array[1] = 'contains'
add % to array[2] just if array[1] = 'contains'

I tried a recursive function but I cannot make it work. Please can anybody help? it's driving me crazy. I tried to solve point 3 like this:

function setFilter( $filtri ){

    $json = json_decode($filtri, false);    

    if(is_array($json)){
        $count_j = count($json);

        if(($count_j==3) && (!is_array($json[0])) && (!is_array($json[1])) && (!is_array($json[2])) ){  

            if($json[1]=='contains'){
              $json[2]='|'.$json[2].'|'; 
            }  

        } else {

            foreach ($json as $item) {                  

                if(is_array($item)){                    
            $item = setFilter( json_encode($item) );    
          };

        }

        }       

  };    

  $json2 = json_encode($json);   
  return $json2;  
}

So the Idea was to put a | where I want to keep the quote, then remove all the quotes, then replace | with quote.

any help? thank you!

****EDIT - I answer my last comment ****

I solved my last problem in the comment like this, what do you think?

// If there are three elements, manipulate the string into the required format
        if (count($json) == 3) {

            if ($json[$i] == 'contains'){
            $json[$i] = 'like';                                     // Change 'contains' to 'like'
            $json[$i + 1] = '"%'.$json[$i + 1].'%"';                // Encapsulate the following token in percentage symbols and quotes
          } else if ($json[$i] == 'notcontains'){
            $json[$i] = 'not like';                                 // Change 'notcontains' to 'not like'
            $json[$i + 1] = '"%'.$json[$i + 1].'%"';                // Encapsulate the following token in percentage symbols and quotes
          } else if ($json[$i] == 'startswith'){
            $json[$i] = 'like';                                    // Change 'startswith' to 'like'
            $json[$i + 1] = '"'.$json[$i + 1].'%"';                // Encapsulate the following token in percentage symbols and quotes
          } else if ($json[$i] == 'endswith'){
            $json[$i] = 'like';                                    // Change 'endswith' to 'like'
            $json[$i + 1] = '"%'.$json[$i + 1].'"';                // Encapsulate the following token in percentage symbols and quotes       
          } else {        
            // I need to find out datatype so that I can add quotes in case of string           
            if(!is_array($json[$i])){                           
              $array = ["=", "<>", "<", ">", "<=", ">="];
              if (stripos(json_encode($array), $json[$i]) !== false) {                
                if (gettype($json[$i]=='string')){
                  $json[$i + 1] = '"'.$json[$i + 1].'"';           // it's a string, add quotes
                }
              };            
            }   
        };

Jack Crawley · Accepted Answer · 2019-11-01 14:53:23Z

2

The best approach is to decode the entire JSON into an array, allowing the function to recursively call itself. The solution below returns the string you are expecting:

// Your input string
$json = '[["Nome","contains","av"],"and",[["IdAreoporto","=",1],"and",["!",["IdAreoporto","=",3]]]]';

function setFilter($filtri) {

    // Decode the JSON string if it is a JSON string, otherwise simply use the array
    $json = (is_array($filtri)) ? $filtri : json_decode($filtri, true);

    // Where the string will be built
    $string = '';

    // For each top-level element in the array
    for ($i = 0; $i < count($json); $i++) {

        // If there are three elements, manipulate the string into the required format
        if (count($json) == 3 && $json[$i] == 'contains') {
            $json[$i] = 'like';                                     // Change 'contains' to 'like'
            $json[$i + 1] = '"%'.$json[$i + 1].'%"';                // Encapsulate the following token in percentage symbols and quotes
        }

        // More string manipulation
        if ($json[$i] == '!') {
            $json[$i] = 'not';
        }

        // If the current token is an array, put it into the function recursively, otherwise simply add it to the string 
        $string .= (is_array($json[$i])) ? setFilter($json[$i]) : ' ' . $json[$i] . ' ' ;
    }

    // Return the cleaned up string back to its caller (will either be another instance of the same function or the original caller)
    return '('.trim($string).')';  
}

echo setFilter($json);

The end result will be:

((Nome like "%av%") and ((IdAreoporto = 1) and (not (IdAreoporto = 3))))

answered Nov 1, 2019 at 14:53

Jack Crawley

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3 Comments

lorife Over a year ago

Hello Jack. I wanted to thank you but rules were I should just click "accept"..but now I have one more question so I want also to thank you so much!! I wanted to ask you this: if you compare the 3rd parameter sometimes it has double quotes and sometimes not (["Nome","contains","av"] ["IdAreoporto","=",1]). so I actually do not need to add quotes always, i need to keep the original quote if present and just add % if the string is contain. Is there a way to keep the original quote? thank you!

lorife Over a year ago

just to explain better, the problem arise when i have "=" as 2nd parameter. in that case the 3rd parameter could be a string or a number. in your answer I cannot see quotes when doing $json[$i + 1] so I don't know if they were present or not.

lorife Over a year ago

I edited my answer with my solution to my last problem. what do you think? (it works!) thanks

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