Question

为什么不给出正确的退出代码？

我的完整代码：

<?php

$numA_m = "2";
$res = '["110","2","1"]';
$numA_s = json_decode($res);
if ($numA_m == 1) {
    $A_num_s = array("1", "2", "3", "4","110");
    $A_nam_s = array("egh", "guide", "irl", "tic", "all");
}
if ($numA_m == 2) {
    $A_num_s = array("1", "2","110");
    $A_nam_s = array("sub", "forg","all");
}
$Rsp = str_replace($A_num_s, $A_nam_s, $numA_s);
$Rsp_In = str_replace($A_nam_s, $A_num_s, $Rsp);
echo '<pre>';
print_r($Rsp);

echo '<p>';

echo '<pre>';
print_r($Rsp_In);
?>

输出应为：

阵列（
      [0] =＆gt;所有
      [1] =＆gt; forg
      [2] =＆gt;子
  ）

阵列（
      [0] =＆gt; 110个
      [1] =＆gt; 2
      [2] =＆gt; 1
  ）

但它是这样的：

阵列
（
      [0] =＆gt; subsub0
      [1] =＆gt; forg
      [2] =＆gt; sub
）阵列
（
      [0] =＆gt; 110个
      [1] =＆gt; 2
      [2] =＆gt; 1
）

我该怎么办？

DEMO： http://phpfiddle.org/main/code/srj-k3d

Answer 1

这会输出所需的结果

$numA_m = "2";
$res = '["110","2","1"]';
$numA_s = json_decode($res);
if ($numA_m == 1) {
    $A_num_s = array("1", "2", "3", "4", "110");
    $A_nam_s = array("egh", "guide", "irl", "tic", "all");
}
if ($numA_m == 2) {
    $A_num_s = array("110", "2", "1"); // changed order
    $A_nam_s = array("all", "forg", "sub");
}
$count = 1;
$Rsp = str_replace($A_num_s, $A_nam_s, $numA_s, $count);
$Rsp_In = str_replace($A_nam_s, $A_num_s, $Rsp, $count);
echo '<pre>';
print_r($Rsp);

echo '<p>';

echo '<pre>';
print_r($Rsp_In);

<强>解释

str_replace按照发送顺序搜索字符串。因此，如果您在搜索1然后110，则会在到达1之前替换110的所有出现次数。

如果您先替换110，它将按预期更换。

Answer 2

首先，str_replace工作正常您可以做的是替换数组中的顺序：

$A_num_s = array("110", "1", "2");
$A_nam_s = array("all", "sub", "forg");

在你的版本中，str_replace将“110”视为两个1（和一个0），并用“sub”（“subsub0”）替换它们。

根据更改后的订单，“110”将首先匹配，导致str_replace按照您提供的顺序搜索您的模式（$A_num_s）。

Answer 3

str_replace按顺序替换，按顺序检查每个替换值的整个替换主题，例如：

http://php.net/manual/en/function.str-replace.php

// Outputs F because A is replaced with B, then B is replaced with C, and so on...
// Finally E is replaced with F, because of left to right replacements.
$search  = array('A', 'B', 'C', 'D', 'E');
$replace = array('B', 'C', 'D', 'E', 'F');
$subject = 'A';
echo str_replace($search, $replace, $subject);

所以，你需要做的就是在1之前替换110，否则首先替换110中的1：

$A_num_s = array("110", "2", "1");

Answer 4

您可以尝试使用array_replace()：或许有点难以理解，但有效;）

$numA_m = "2";
$res = '["110","2","1"]';
$numA_s = json_decode($res);
if ($numA_m == 1) {
    $A_num_s = array("1", "2", "3", "4","110");
    $A_nam_s = array("egh", "guide", "irl", "tic", "all");
}
if ($numA_m == 2) {
    $A_num_s = array("1", "2","110");
    $A_nam_s = array("sub", "forg","all");
}

$Rsp = array_replace($numA_s, $A_num_s, array_reverse($A_nam_s));
$Rsp_In = array_replace($Rsp, $A_nam_s, array_reverse($A_num_s));

echo '<pre>';
print_r($Rsp);


echo '<p>';

echo '<pre>';
print_r($Rsp_In);
?>

str_replace无法正常工作

4 个答案: