使用php轉換為json

Question

我有一個電子表格，我正在將其轉換為json。我可以使用php代碼進行轉換。但我想為數組命名。我該怎么做..請幫助..Php和輸出以及所需的輸出被提及。

 Required output ["Name"{"Timestamp":"7\\/24\\/2015 12:42:41","Name":"ADADSADS","Type":"ASDSD","Place":"ASDSADD","Date":"ASDSD","Time":"ASDSD","Free":"ASDSD","Organizer":"ASDSD","Contact":"ASDSD","Description":"ASDSD","id":0}] 

 Output from the below code [{"Timestamp":"7\\/24\\/2015 12:42:41","Name":"ADADSADS","Type":"ASDSD","Place":"ASDSADD","Date":"ASDSD","Time":"ASDSD","Free":"ASDSD","Organizer":"ASDSD","Contact":"ASDSD","Description":"ASDSD","id":0}] 

 <?php /* * Converts CSV to JSON * Example uses Google Spreadsheet CSV feed * csvToArray function I think I found on php.net */ header('Content-type: application/json'); // Set your CSV feed $feed = 'google doc url'; // Arrays we'll use later $keys = array(); $newArray = array(); // Function to convert CSV into associative array function csvToArray($file, $delimiter) { if (($handle = fopen($file, 'r')) !== FALSE) { $i = 0; while (($lineArray = fgetcsv($handle, 4000, $delimiter, '"')) !== FALSE) { for ($j = 0; $j < count($lineArray); $j++) { $arr[$i][$j] = $lineArray[$j]; } $i++; } fclose($handle); } return $arr; } // Do it $data = csvToArray($feed, ','); // Set number of elements (minus 1 because we shift off the first row) $count = count($data) - 1; //Use first row for names $labels = array_shift($data); foreach ($labels as $label) { $keys[] = $label; } // Add Ids, just in case we want them later $keys[] = 'id'; for ($i = 0; $i < $count; $i++) { $data[$i][] = $i; } // Bring it all together for ($j = 0; $j < $count; $j++) { $d = array_combine($keys, $data[$j]); $newArray[$j] = $d; } // Print it out as JSON echo json_encode($newArray); ?> 

Answer 1

那是無效的json。 這樣做的目的是什么？

substr_replace（）用替換中給定的字符串替換由start和（可選）length參數分隔的字符串副本。

混合substr_replace（混合$ string，混合$ replacement，混合$ start [，混合$ length]）

echo substr_replace(json_encode($newArray), '"Name"', 1, 0);

但是，如果您的意思是{ "Name": <JSON> }則可以執行以下操作：

echo json_encode(array("Name" => $newArray));

Answer 2

在此行之前echo json_encode($newArray);

將數據分配給數組鍵..像這樣

$newArray2['name']=$newArray;
echo json_encode($newArray2);