Question

我有2个php文件。第一个php：

 //store values:
    $Hitcount=0;
    $result=array();
    while statement {
    $col = 0;
    $result[$Hitcount][$col]=$hit_name;
    $result[$Hitcount][++$col]=$hit_link;
    $Hitcount++;
    }
    //check stored values
    foreach ($result as $key => $value) 
    {
      foreach ($value as $k => $v) 
       {
            echo '$v'; echo",";
       }
    }

我从之前的回声中得到以下信息：

gi|1786181|gb|AE000111|ECAE000111,http://www.ncbi.nlm.nih.gov/nuccore/1786181?report=genbank,
gi|1786250|gb|AE000117|ECAE000117,http://www.ncbi.nlm.nih.gov/nuccore/1786250?report=genbank,
gi|1786240|gb|AE000116|ECAE000116,http://www.ncbi.nlm.nih.gov/nuccore/1786240?report=genbank, gi|1786217|gb|AE000114|ECAE000114,http://www.ncbi.nlm.nih.gov/nuccore/1786217?report=genbank,

现在，我想将$ result数组发布到另一个php文件中。要发布，第一个php包含：

foreach ($result as $key => $value) 
                {
                    foreach ($value as $k => $v) 
                    {
                        echo "<input type='hidden' name='result[$k]' value='$v'/>";
                    }
                }

接收$ result数组，第二个php：

$result = empty($_POST['result']) ? array() : $_POST['result'];
foreach ($result as $key => $value ) 
                {
                    foreach ($value as $k => $v) 
                    {
                        echo $v;
                    }
                }

但是，回声给了我：

Invalid argument supplied for foreach().

这意味着第二个foreach（）声明。

请帮忙吗？

编辑1： 它现在通过修复第一个PHP作为：

foreach ($result as $key => $value) 
{
    foreach ($value as $k => $v) 
    {
        echo "<input type='hidden' name='result[$key][$k]' value='$v'/>";
                                               ^^^^^^ add second key here
    }
}

现在，如何在数据库中保存$ result？我想：

insert row1 in the database and put $result[0][0] in col1 and put $result[0][1] in col2
insert row2 in the database and put $result[1][0] in col1 and put $result[1][1] in col2
insert row3 in the database and put $result[2][0] in col1 and put $result[2][1] in col2
and so on.

现在我尝试：

 foreach ($result as $key => $value) 
                {
                    foreach ($value as $k => $v) 
                    {
                        $sql = "INSERT INTO BlastResultFact (Hit) VALUES ('$v')";
        $stmt = sqlsrv_query( $conn, $sql);
        if( $stmt === false ) {
             die( print_r( sqlsrv_errors(), true));
                    }
                }}

it insert row1 in the database and put $result[0][0] in col1 and
insert row2 in the database and put $result[0][1] in col1 and 
insert row3 in the database and put $result[1][0] in col1 and so on.

我相信我必须在插入声明中包含col2（HitLink），但是$ v呢？

$sql = "INSERT INTO BlastResultFact (Hit, HitLink) VALUES ('$v')";

Answer 1

从$_POST['result']获取的数组是字符串而不是数组。您需要explode()这样的字符串：

$result = empty($_POST['result']) ? array() : explode ( '|' ,$_POST['result']);

Answer 2

问题是您的input代码是1维的，因此您的内部foreach将失败，因为$_POST['result']的每个元素都是字符串，而不是数组

$result = empty($_POST['result']) ? array() : $_POST['result'];
foreach ($result as $key => $value ) 
{
    // $value is a string!
    foreach ($value as $k => $v) 
    {
        echo $v;
    }
}

要解决此问题，请修改input代码以生成二维数组：

foreach ($result as $key => $value) 
{
    foreach ($value as $k => $v) 
    {
        echo "<input type='hidden' name='result[$key][$k]' value='$v'/>";
                                               ^^^^^^ add second key here
    }
}

现在你应该在$_POST['result']中获得一个2d数组，并能够使用原始的嵌套循环迭代数组

Answer 3

您可以使用此方法进行游戏。

foreach($array as $index=>$val) 
{
    //suppose you have three field
    $itemid=$_POST['item'][$index]['value'];
    $quantity=$_POST['quantity'][$index]['value'];
    $price=$_POST['unitprice'][$index]['value'];
}

如何回应从php发布的multidimentional数组

3 个答案: