如何通过运行某种脚本将这7列精简为5列。
我记得你可以使用regex/php来做这件事，但如果我能回忆起我们是如何做的，那就麻烦了
示例代码（GEO IP数据库（115000条线路）
"3231296768","3231297023","ripencc","702518400","EU","EU","European Union"
"3231297024","3231297279","ripencc","441763200","EU","EU","European Union"
"3231297280","3231297535","ripencc","702518400","EU","EU","European Union"
"3231297536","3231297791","ripencc","702518400","EU","EU","European Union"
"3231297792","3231298047","ripencc","702518400","EU","EU","European Union"
"3231298048","3231298303","ripencc","702518400","EU","EU","European Union"
"3231298304","3231298559","ripencc","702518400","EU","EU","European Union"
我需要从每行中删除第3列和第4列。
感谢您的帮助。
虽然jimw的答案通常是最好的答案，但如果你想要一个纯PHP解决方案，我建议如下：
$input = 'input.txt';
$output = 'output.txt';
if (false !== ($ih = fopen($input, 'r'))) {
$oh = fopen($output, 'w');
while (false !== ($data = fgetcsv($ih))) {
// this is where you build your new row
$outputData = array($data[0], $data[1], $data[4], $data[5], $data[6]);
fputcsv($oh, $outputData);
}
fclose($ih);
fclose($oh);
}
根据"排序脚本"answers"regex/PHP"，我推断您只想完成这项工作，而不在乎使用什么语言。如果您使用*nix:
cut -d, -f1,2,5,6,7 file.csv
"cut"是一个标准的unix命令行实用程序，可以在从OSX到AIX的所有应用程序中找到。我使用的论点是：
-d, # this sets the 'delimiter' to a comma, for CSV
-f1,2,5,6,7 # this selects which fields to print
因此，它合并成一个文件，其中每一行都由逗号分隔的字段组成，并打印出其中的一到五个字段
在任何编程语言中都可以达到同样的效果。我对PHP不是很了解，所以我不会尝试用PHP制作它。
编辑：来自PHP文档，略有调整：
function apply_quotes($string) {
return '"'.$string.'"';
}
$row = 1;
if (($handle = fopen("test.csv", "r")) !== FALSE) {
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$data = array_map("apply_quotes", $data);
echo join(",", array($data[0], $data[1], $data[4], $data[5], $data[6]))."'n";
}
fclose($handle);
}