Name, Id, Address, Place,
John, 12, "12 mark street", "New York",
Jane, 11, "11 bark street", "New York"...
我有大约500个coloumns.我想将其转换为JSON,但我希望输出看起来像:
{
"name": [
"John",
"Jane"
],
"Id": [
12,
11
],
"Address": [
"12 mark street",
"12 bark street"
],
"Place": [
"New York",
"New York"
]
}
使用PHP,我如何迭代CSV文件,以便我可以使第一行中的每一列成为一个数组,该数组在所有其他行的同一列中保存值?
解决方法:
这将是一个通用方法,对任何命名列的amoutn都有效.
如果它们是静态的,则直接解决它们会更短
<?
$result = array();
if (($handle = fopen("file.csv", "r")) !== FALSE) {
$column_headers = fgetcsv($handle); // read the row.
foreach($column_headers as $header) {
$result[$header] = array();
}
while (($data = fgetcsv($handle)) !== FALSE) {
$i = 0;
foreach($result as &$column) {
$column[] = $data[$i++];
}
}
fclose($handle);
}
$json = json_encode($result);
echo $json;