使用php编码json吗? [英] encode json using php?
问题描述
我想使用php编码功能获取json,如下所示
I want to get json with php encode function like the following
<?php
require "../classes/database.php";
$database = new database();
header("content-type: application/json");
$result = $database->get_by_name($_POST['q']); //$_POST['searchValue']
echo '{"results":[';
if($result)
{
$i = 1;
while($row = mysql_fetch_array($result))
{
if(count($row) > 1)
{
echo json_encode(array('id'=>$i, 'name' => $row['name']));
echo ",";
}
else
{
echo json_encode(array('id'=>$i, 'name' => $row['name']));
}
$i++;
}
}
else
{
$value = "FALSE";
echo json_encode(array('id'=>1, 'name' => "")); // output the json code
}
echo "]}";
我希望输出json是类似的
i want the output json to be something like that
{"results":[{"id":1,"name":"name1"},{"id":2,"name":"name2"}]}
,但输出json如下所示
but the output json is look like the following
{"results":[{"id":1,"name":"name1"},{"id":2,"name":"name2"},]}
当您意识到结尾处有逗号时,我想将其删除,因此它可以是正确的json语法,如果在有多个结果的情况下删除了echo ",";
,则json会像这样生成{"results":[{"id":1,"name":"name1"}{"id":2,"name":"name2"}]}
,并且语法也是错误的
As you realize that there is comma at the end, i want to remove it so it can be right json syntax, if i removed the echo ",";
when there's more than one result the json will generate like this {"results":[{"id":1,"name":"name1"}{"id":2,"name":"name2"}]}
and that syntax is wrong too
希望每个人都明白我的意思,任何想法都会受到赞赏
Hope that everybody got what i mean here, any ideas would be appreciated
推荐答案
如果我是你,我不会json_encode
每个单独的数组,而是将这些数组合并在一起,然后在最后将json_encode
合并的数组合并.以下是使用 5.4的短数组语法的示例:
If I were you, I would not json_encode
each individual array, but merge the arrays together and then json_encode
the merged array at the end. Below is an example using 5.4's short array syntax:
$out = [];
while(...) {
$out[] = [ 'id' => $i, 'name' => $row['name'] ];
}
echo json_encode($out);
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