Question

所以我对每个循环都有一些，并且我试图在最后回显$ sql的值。我实际上是在PDO中使用它，但回声显示了这个问题。

$values = "1,2";
$values = explode(',',$values);

$set1 = "2,4";
$set1 = explode(',',$set1);
$set2 = "3,2";
$set2 = explode(',',$set2);

foreach($set1 as $set1val){
  if ($set1val==2) {
    $sql = "sql one runs";//should run on 1st iteration since 2 will equal 2
    echo 'hi';//should echo on 1st iteration since 2 will equal 2
  }
}

foreach($set2 as $set2val){
  if ($set2val==2) {
    $sql = "sql two runs";//should run on 2nd iteration since 2 will equal 2
    echo 'bye';//should echo on 2nd iteration since 2 will equal 2
  }
}

foreach($values as $value){
  echo $sql;
  $stmt = $db->query($sql);
}

//The Result Output
'hi'  'bye'  'sql two runs'  'sql two runs' <-- doesn't echo 'sql one runs' 

//The Output Required
'hi'  'bye'  'sql one runs'  'sql two runs' <-- Works great, but not outputting this

如何输出所需的输出？知道如何做到这一点，以便正确的声明回显/运行？

Answer 1

它永远不会回显sql one runs，因为第二个foreach会用sql two runs覆盖它。如果需要在脚本末尾保存sql字符串，可以将它们存储在数组中。很难指定如何不清楚$values与SQL的关系，即哪个SQL应与哪个$value一起运行。

Answer 2

问题是，您首先将sql one runs保存在$sql中，但您使用$sql做的下一件事是保存sql two runs，覆盖之前的值。因此，您需要将这些值保存在数组中，并修改最后一个foreach。像这样：

foreach($set1 as $set1val){
  if ($set1val==2) {
    $sql[0] = "sql one runs";//should run on 1st iteration since 2 will equal 2
    echo 'hi';//should echo on 1st iteration since 2 will equal 2
  }
}

foreach($set2 as $set2val){
  if ($set2val==2) {
    $sql[1] = "sql two runs";//should run on 2nd iteration since 2 will equal 2
    echo 'bye';//should echo on 2nd iteration since 2 will equal 2
  }
}

foreach($sql as $value){
  echo $value;
  $stmt = $db->query($value);
}

PDO和For Each循环：输出不正确的变量

2 个答案: