所以我有以下内容:
$audio1_1 = "x";
$audio2_1 = "z";
$audio3_1 = "r";
$audio4_1 = "b";
$audio5_1 = "x";
$audio6_1 = "z";
$audio7_1 = "r";
$audio8_1 = "b";
for ($x = 1; $x <= 8; $x++) {
//// what to put here to access the corresponding value???
}
我的意思是当$ x = 1时我想获得$ audio1_1的值,当$ x = 2时我想获得$ audio2_1的值,依此类推。 谢谢!
答案 0 :(得分:4)
如果你想存储许多值,你应该使用数组
$audio[] = "x";
$audio[] = "z";
$audio[] = "r";
$audio[] = "b";
$audio[] = "x";
$audio[] = "z";
$audio[] = "r";
$audio[] = "b";
foreach ($audio as $value) {
//// do something with $value
}
如果你不能这样做,你必须使用你的$ audioX_1 var名称,你可以做这样的事情
for ($x = 1; $x <= 8; $x++) {
$varname = "audio{$x}_1";
$value = $$varname;
}
答案 1 :(得分:0)
只需使用数组。
$audio = array("x", "z", "r", "b", "x", "z", "r", "b");
foreach ($audio as $audioTemp)
{
$audioTemp //will be x then z then r and so on
}
答案 2 :(得分:0)
$audio1_1 = "x";
$audio2_1 = "z";
$audio3_1 = "r";
$audio4_1 = "b";
$audio5_1 = "x";
$audio6_1 = "z";
$audio7_1 = "r";
$audio8_1 = "b";
for ($x = 1; $x <= 8; $x++) {
$varName="audio".$x."_1";
echo $$varName;
}
答案 3 :(得分:0)
$audio1_1 = "x";
$audio2_1 = "z";
$audio3_1 = "r";
$audio4_1 = "b";
$audio5_1 = "x";
$audio6_1 = "z";
$audio7_1 = "r";
$audio8_1 = "b";
for ($x = 1; $x <= 8; $x++) {
$var = "audio".$x."_1";
print $$var;
}