Question

我试图通过for循环运行一个数组，将检查的项目附加到每个数组上，每个数组附加后缀为$ n（1-3）。谢谢你的建议我更近了一步，现在我得到一个项目的每个后缀。我希望每个人都在数组中作为自己的项目。任何人都可以看到错误吗？

我已经更新了我的代码，它离解决方案更近了一步，它正是我的目标。

$equip = 'Phone';
$ailments_checkvar =  explode(', ', 'Cracked, Scratched, Bent, Twisted');

foreach ($ailments_checkvar as &$value) {
    $value = 'directory/'.$equip.'_'.$value.'';
}
unset($value);
$duplicateArray = $ailments_checkvar;

foreach ($ailments_checkvar as $key) {
    $duplicateArray[] = $key;
}

foreach ($ailments_checkvar as $key) {
    $duplicateArray[] = $key;
}
for ($n = 1; $n <= 3; $n++) {

foreach ($duplicateArray as &$valueN) {
    $valueN = $valueN.'_0'.$n.'.pdf';
}
}
unset($valueN);
print_r ($duplicateArray);

获取此

Array ( [0] => directory/Phone_Cracked_01.pdf_02.pdf_03.pdf 
[1] => directory/Phone_Scratched_01.pdf_02.pdf_03.pdf 
[2] => directory/Phone_Bent_01.pdf_02.pdf_03.pdf 
[3] => directory/Phone_Twisted_01.pdf_02.pdf_03.pdf 
[4] => directory/Phone_Cracked_01.pdf_02.pdf_03.pdf 
[5] => directory/Phone_Scratched_01.pdf_02.pdf_03.pdf 
[6] => directory/Phone_Bent_01.pdf_02.pdf_03.pdf 
[7] => directory/Phone_Twisted_01.pdf_02.pdf_03.pdf 
[8] => directory/Phone_Cracked_01.pdf_02.pdf_03.pdf 
[9] => directory/Phone_Scratched_01.pdf_02.pdf_03.pdf 
[10] => directory/Phone_Bent_01.pdf_02.pdf_03.pdf 
[11] => directory/Phone_Twisted_01.pdf_02.pdf_03.pdf 
)

想要做到这一点......

Array ( 
[0] => directory/Phone_Cracked_01.pdf 
[1] => directory/Phone_Cracked_02.pdf 
[2] => directory/Phone_Cracked_03.pdf 
[3] => directory/Phone_Scratched_01.pdf 
[4] => directory/Phone_Scratched_02.pdf
[5] => directory/Phone_Scratched_03.pdf
[6] => directory/Phone_Bent_01.pdf 
[7] => directory/Phone_Bent_02.pdf 
[8] => directory/Phone_Bent_03.pdf 
[9] => directory/Phone_Twisted_01.pdf
[10] => directory/Phone_Twisted_02.pdf
[11] => directory/Phone_Twisted_03.pdf 
)

Answer 1

请看这里的陈述，

$valueN = $ailments_checkvar.'_0'.$n.'.pdf';
          ^^^^^^^^^^^^^^^^^^

$ailments_checkvar是一个数组，而不是一个字符串。在上面的语句中，您尝试将数组转换为字符串。它应该是，

$valueN = $valueN.'_0'.$n.'.pdf';

更新（1）：

根据您在上面发布的要求，解决方案将是这样的：

$equip = 'Phone';
$ailments_checkvar =  explode(', ', 'Cracked, Scratched, Bent, Twisted');

$path_array = array();
foreach($ailments_checkvar as $property_status){
    for($i = 1; $i <= 3; ++$i){
        $path_array[] = 'directory/' . $equip . '_' . $property_status . '_0' . $i . '.pdf';
    }
}
echo '<pre>'; print_r ($path_array);

这里是live demo。

Answer 2

for ($z=1;$z<=count($ailments_checkvar);$z++) {
$ailments_checkvar[$z-1] = $ailments_checkvar[$z-1].'_0'.$z.'.pdf';
}
unset($z);

你曾经做过什么导致$ ailments_checkvar的堆栈问题请查看这是优化代码。

你的错误在于$ valueN = $ ailments_checkvar .'_ 0'。$ n。'。pdf';

将每个项目附加到具有多个后缀的数组中，然后创建新数组

2 个答案:

更新（1）：