Biw*_*ong -4 php arrays url replace
我有一个.csv文件,我已经转换成一个数组$outputData.$data[6]每个数组的第6个元素包含一个图像文件路径,例如
http://images.pleaserusa.com/pic/unique.image.jpg
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我需要删除http://images.pleaserusa.com/pic/每个图像的文件路径,只留下图像名称本身,例如
unique.image.jpg
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我已经看过这里和谷歌并试图使用preg_match()和str*功能,但我真的没有得到它.
Rizer123我已经放置了"basename"函数,但无法在元素内部替换它.我只能让它在数组之外回显/打印/我已经放置了下面的代码,下面是一个摘录数组.
domina-456-b.jpgArray
(
[0] => DOM456/B
[1] => 6" Lace-Up Pump W/ D-Ring&Ribbon Lace
[2] => Devious
[3] => Single Soles
[4] => DOMINA-456
[5] => Blk Pat
[6] => http://images.pleaserusa.com/pic/domina-456-b.jpg
[7] => 5-15
)
domina-460-b.jpgArray
(
[0] => DOM460/B
[1] => 6" Oxford Lace Up Pump
[2] => Devious
[3] => Single Soles
[4] => DOMINA-460
[5] => Blk Pat
[6] => http://images.pleaserusa.com/pic/domina-460-b.jpg
[7] => 5-16
)
<?php
$input = 'Parser/Inv_item.csv';
$output = 'hd_inv_items.csv';
if (false !== ($ih = fopen($input, 'r'))) {
$oh = fopen($output, 'w');
header('Content-Type: text/plain');
while (false !== ($data = fgetcsv($ih))) {
// this is where you build your new row
$outputData = array($data[0], $data[1], $data[2], $data[3], $data[5],$data[6], $data[9], $data[8]);
$path = $data[9];
$name = basename($path); // $name == '1.jpg'
print_r($name);
print_r($outputData);
//fputcsv($oh, $outputData);
}
fclose($ih);
fclose($oh);
}
?>
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如您所见,图像文件路径仍在数组中,并且basename结果在数组外部打印.我该如何解决这个问题
关心Biwwabong
不需要爆炸或字符串操作,简单basename()就足够了:
$path = 'http://images.pleaserusa.com/pic/unique.image.jpg';
echo basename($path);
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输出:
unique.image.jpg
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pathinfo() 会工作:
echo pathinfo($path, PATHINFO_BASENAME);
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