删除PHP数组中用于JSON格式的逗号

我正在输出一个json格式的数组。如果地址字段和市场字段之间用逗号分隔，那么如果字段不存在，我想删除逗号，这样我就不会得到带有逗号的空格。我曾尝试过内爆，但这似乎不起作用。。。正确的做法是什么

  $data = array(
    'ID' => $Member->memberID(),
    'Name' => $Member->first_name() . ' ' . $Member->last_name(),
    'Job Title' => $Member->expert_job_title(),
    'Organisation' => $Member->expert_org_name(),
    'Organisation Type' => $Member->expert_org_type(),
    'Website' => 'http://' . $Member->expert_org_website(),
    'Phone' => $Member->expert_org_phone(),
    'Expertise' => $Member->expert_org_desc(),
    'Markets' => implode(', ', array($Member->expert_org_market_medical(), $Member->expert_org_market_pharmaceuticals(), $Member->expert_org_market_agriculture(), $Member->expert_org_market_food(), )),
    'Address' => $Member->expert_address_one() . ', ' . $Member->expert_address_two() . ', ' . $Member->expert_address_town_city() . ', ' . $Member->expert_address_county_state() . ', ' . $Member->expert_org_country() . ', ' . $Member->expert_address_code(),
    'Latitude' => $Member->expert_org_latitude(),
    'Longitude' => $Member->expert_org_longitude()
  );



header('Content-Type: application/json');

echo json_encode($data, JSON_PRETTY_PRINT);

用答案更新

'Markets' => implode(', ', array_filter(array($Member->expert_org_market_medical(), $Member->expert_org_market_pharmaceuticals(), $Member->expert_org_market_agriculture(), $Member->expert_org_market_food(), ))),
'Address' => implode(', ', array_filter(array($Member->expert_address_one(), $Member->expert_address_two(), $Member->expert_address_town_city(), $Member->expert_address_county_state(), $Member->expert_org_country(), $Member->expert_address_code(), ))),

像这样用于地址和市场字段

<?php
echo "without array filter = " . implode(', ', array('a', '', 'b', '', ));
echo "<br/>";
echo "with array filter = ". implode(', ', array_filter(array('a', '', 'b', '', )));
?>

上述示例将输出：

Array
(
    [0] => foo
    [2] => -1
)

---

有一些你做的事情。但是，您可以尝试使用

数组过滤器

也许用preg_替换

 $json = json_encode($data);
 $json = preg_replace('/(^, )|( ,)/mi', '', $json);
 $data = json_decode($json);

快速而肮脏的方法是修剪任何看起来像“，”替换为“，”的字符串，我认为你可以用空间内爆（“”，…）
$Markets = array($Member->expert_org_market_medical(), $Member->expert_org_market_pharmaceuticals(), $Member->expert_org_market_agriculture(),  $Member->expert_org_market_food());
$Markets = array_filter($Markets, function($item) {
    // return if that strings isn't null
    // maybe you want to add another restrictions here
    return strlen($item);
}, $Markets);
// all empty items was removed.|
$Markets = implode(',', $Markets);

 $json = json_encode($data);
 $json = preg_replace('/(^, )|( ,)/mi', '', $json);
 $data = json_decode($json);