使用php转换为json

Question

我有一个正在转换为json.I的电子表格，我可以使用php code.But进行转换，我想将数组命名为.How，我可以做这个..Please help..Php，并提到了输出和所需的输出。

​

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

在这行之前echo json_encode($newArray);

将数据分配给数组键。像这样

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

Answer 2

那是无效的，json。这样做有什么意义？

substr_replace()用替换中给出的字符串替换由开始参数和(可选)长度参数分隔的字符串副本。 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);

但如果你指的是{ "Name": <JSON> }，那么你可以：

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