Question

我有我的PHP脚本的麻烦从值内修剪空白，因为我输出PostgreSQL的查询结果。的注释行是从print_r();

<?php 

$db = pg_connect("host=localhost dbname=test user=test password=test"); 
$result = pg_query($db, "SELECT * FROM table WHERE field = 'test'"); 
$array = pg_fetch_all($result); 

//Array ([0] => Array ([id_number] => 214 [country] => Zanzibar [sector] => Unguja [site] => Chumbe Island Coral Park (CHICOP) [username] => n) [1] => Array ([id_number] => 213 [country] => Zanzibar [sector] => Unguja [site] => Chumbe Island Coral Park (CHICOP) [username] => n) 

//************* 
$array_explode = explode(",", $array); 
$array_trimmed = array_map("trim", $array_explode); 
$array_implode = implode(",", $array_trimmed); 
unset($array_explode); 
//************* 

$fp = fopen('file.csv', 'w'); 

foreach ($array_implode as $fields) { 

fputcsv($fp, $fields); 
} 
fclose($fp); 


?> 

返回的数组如果删除了分则CSV输出（只是一大堆不必要的空白的，我已经都具有和不具有implode()尝试之间的行。此。一直在推动我逼疯了和任何帮助将不胜感激

Answer 1

我能得到它使用您的代码工作：

<?php 
$array = Array (0 => Array ('id_number' => 214, 'country' => 'Zanzibar', 'sector' => 'Unguja', 'site' => 'Chumbe Island Coral Park (CHICOP)', 'username' => 'n'), 
     1 => Array ('id_number' => 213, 'country' => 'Zanzibar', 'sector' => 'Unguja', 'site' => 'Chumbe Island Coral Park (CHICOP)', 'username' => 'n')); 

$fp = fopen('file.csv', 'w'); 
foreach ($array as $fields) { 
     $array_trimmed = array_map("trim", $fields); 
     fputcsv($fp, $array_trimmed); 
} 
fclose($fp); 

从FILE.CSV输出：

214,Zanzibar,Unguja,"Chumbe Island Coral Park (CHICOP)",n 
213,Zanzibar,Unguja,"Chumbe Island Coral Park (CHICOP)",n