如何将json格式的复选框数据转换为列表？

问题描述：

How would I convert a json string that contains checkbox values to a list?

For example, I have the following json string : {"User1":"Checked","User2":"","User3":"Checked"}

I want to pull a list from the above data, and assign it to a variable. The list should of course, only output the checked elements as such :

User1 User3

and ignore the unchecked items.

Is this possible?

Original thought was to do this in a jqgrid layout, original question here : https://stackoverflow.com/questions/34615244/best-way-to-do-jqgrid-with-json-strings

Full Array Code(processing side) :

$status = $data['data-invoice-status']; // this is an array
                $status_array = array(); // create new array
                $status_names = array(1 => "Service Call",2 => "Estimate",3 => "Bid Accepted",4 => "Unstaged",5 => "Staged",6 => "Assigned",7 => "In Progress",8 => "Job Complete",9 => "Billed");
                for($i = 1; $i <= 9; $i++){ // loop from 1 to 5
                if(in_array($i, $status)){ // if value exists (has been selected), stack 'checked', if not, stack ''.
                    $status_array[$status_names[$i]] = "checked";
                } else {
                    $status_array[$status_names[$i]] = "";
                }
                }
                $status_json = mysqli_real_escape_string($this->_con, json_encode($status_array)); // encode the array into JSON and then escape it.

I then save $statusd into my db whenever anyone updates the form, but no data is transferred.

答

Decode the JSON to an array, then get the keys where the value equals 'Checked':

$list = array_keys(json_decode($json, true), 'Checked');

Or possibly to filter out empty items instead of searching for 'Checked', but then you still have keys instead of values:

$list = array_filter(json_decode($json, true));

Then implode() or foreach() or whatever to work with the array.

Based on your edits:

You really should use a related table and save each user to a new row, but
You're encoding then adding slashes mysqli_real_escape_string and then trying to decode it again

Assuming you're not going to do #1 then just this:

$statusd = mysqli_real_escape_string($this->_con,
                                     implode(",",  array_keys($status_array, 'Checked')));

如何将json格式的复选框数据转换为列表？

Based on your edits:

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