我正在用PHP将数组数据保存到.json文件中。.json文件然后由一些JS在其他地方调用，并放入一个变量中。
以下是PHP的代码：
$fetch = mysqli_query($connection, "SELECT * FROM mrb_prprty WHERE `nationalid` = '". $nationalid ."' AND `author` = '" .$username. "'");
$return_arr = array();
while ($row = mysqli_fetch_array($fetch)) {
$row_array['firstName'] = $row['fname'];
$row_array['lastName'] = $row['lname'];
$row_array['nationality'] = $row['nationality'];
$row_array['phone'] = $row['mobile'];
$row_array['email'] = $row['email'];
$row_array['street'] = $row['street'];
$row_array['towncity'] = $row['towncity'];
$row_array['postcode']= $row['postcode'];
$row_array['county'] = $row['county'];
$row_array['country'] = $row['country'];
$row_array['rentpw'] = $row['rentpw'];
$row_array['nationalid'] = $row['nationalid'];
$row_array['passport'] = $row['passport'];
$row_array['isFavorite'] = $row['favourite'];
array_push($return_arr,$row_array);
}
$json_mrb = json_encode($return_arr);
$myFile = "userdata/". $username .".json";
$method = (file_exists($myFile)) ? 'a' : 'w';
$fh = fopen($myFile,$method);
fwrite($fh, $json_mrb."'n");
当这是第一次执行时，它将检测文件是否存在&根据结果创建一个新文件或附加到现有文件。
以下是第一次运行上面的PHP代码后的".json"文件预览：
[{"firstName":"Ryan","lastName":"Butterworth","nationality":"rKqF01JNdAB1'/EjPW1sulcHzA+Lqp3y'/aaNtpixFphA=","phone":"6jAbOoIal2Eh73neklZgXcA2jJ3831uRiJWE74fADD8=","email":"gpLkiFg4ju4jv4QK6nftnPCqmzdENbsDt4ayI4L1rrkXvg0VJkxmP9Ac57fs4Ar'/","street":"22 Newton Drive","towncity":"Blackpool","postcode":"FY38BS","county":"England","country":"England","rentpw":"VCXaMj'/V6YNExbREuI08eOuINutZRxcx4vv5s2C2FR4=","nationalid":"dNDgHnJJg4rFgbb6qOEC'/dVlnJUNWLCZASo9X8GP5EE=","passport":"123456","isFavorite":""}]
我遇到的问题是，当您第二次运行PHP代码时，.json文件将如下所示：
[{"firstName":"Ryan","lastName":"Butterworth","nationality":"rKqF01JNdAB1'/EjPW1sulcHzA+Lqp3y'/aaNtpixFphA=","phone":"6jAbOoIal2Eh73neklZgXcA2jJ3831uRiJWE74fADD8=","email":"gpLkiFg4ju4jv4QK6nftnPCqmzdENbsDt4ayI4L1rrkXvg0VJkxmP9Ac57fs4Ar'/","street":"22 Newton Drive","towncity":"Blackpool","postcode":"FY38BS","county":"England","country":"England","rentpw":"VCXaMj'/V6YNExbREuI08eOuINutZRxcx4vv5s2C2FR4=","nationalid":"dNDgHnJJg4rFgbb6qOEC'/dVlnJUNWLCZASo9X8GP5EE=","passport":"123456","isFavorite":""}]
[{"firstName":"Ryan","lastName":"Butterworth","nationality":"rKqF01JNdAB1'/EjPW1sulcHzA+Lqp3y'/aaNtpixFphA=","phone":"6jAbOoIal2Eh73neklZgXcA2jJ3831uRiJWE74fADD8=","email":"gpLkiFg4ju4jv4QK6nftnPCqmzdENbsDt4ayI4L1rrkXvg0VJkxmP9Ac57fs4Ar'/","street":"22 Newton Drive","towncity":"Blackpool","postcode":"FY38BS","county":"England","country":"England","rentpw":"VCXaMj'/V6YNExbREuI08eOuINutZRxcx4vv5s2C2FR4=","nationalid":"dNDgHnJJg4rFgbb6qOEC'/dVlnJUNWLCZASo9X8GP5EE=","passport":"123456","isFavorite":""}]
Javascript将".json"文件加载为一个变量，如：var contacts = //fileoutputhere，但正如您所看到的，.json文件现在有两个数组列表，因此它会遇到错误。
不要像这样保存到文件中：[{ ...info... }] [{ ...info... }]
我希望它像这样保存：[{ ...info... },{ ...info... }]
我知道我必须以某种方式调整PHP，但不确定如何做到这一点。有什么建议吗？
使用您的方法，目标文件中有多个JSON字符串，而不是一个唯一的JSON。
以这种方式启动代码：
$myFile = "userdata/". $username .".json";
if( file_exists( $myFile ) )
{
$data = file_get_contents( $myFile );
$return_arr = json_decode( $data, TRUE );
}
else
{
$return_arr = array();
}
此时，如果文件存在，则在$return_arr中，您之前已保存了值。我使用json_encode的第二个参数来返回一个关联数组，以便与新添加的值保持一致，但在您的情况下，它是完全可选的。如果该文件不存在，则$return_arr是一个空数组。
然后继续您的代码：
$fetch = mysqli_query($connection, "SELECT * FROM mrb_prprty WHERE `nationalid` = '". $nationalid ."' AND `author` = '" .$username. "'");
while ($row = mysqli_fetch_array($fetch)) {
(...)
}
$json_mrb = json_encode($return_arr);
并覆盖现有文件（或创建一个新文件）：
$fh = fopen( $myFile, 'w' );
fwrite( $fh, $json_mrb."'n" );
您需要将数据附加到json数组中，因此必须首先读取文件。这里有一些伪代码让你开始：
var myFile = "some-file.json";
var return_arr = array();
if(fileExists($myFile)){
//maybe add some error handling, in case the file contents are not json?
$return_arr = json_decode(get_file_contents($myFile));
}
while($row = getRows()){
$row_array[x] = ...
array_push($return_arr, $row_array);
}
$fh = $fopen($myFile, 'wa+');
fwrite(json_encode($return_arr));