php - $_POST 中的问题

Question

我得到了我的问题 :D 它是关于 $_POST 使用我的 id 无法在插入提交按钮中访问我插入隐藏在我的 id 的 if 语句中

if(array_key_exists('myid',$_POST))
{

$query1="select * from patient where id=".$_POST['myid'];
$result1=mysql_query($query1);
echo $query1;
//var_dump($_POST['myid']);
$num2=Mysql_num_rows($result1);
$num3=Mysql_num_fields($result1);
//clinical file ro neshun mide
        if($num2>0)
        {
        echo "<table border=2>";
        for($i=0;$i<$num2;$i++)
        {
        $row=mysql_fetch_row($result1);


        echo"<td>id</td><td>name</td><td>Lastname</td><td>Info</td><td>Sympthoms</td><td>Diagnosis</td>";
        echo "<tr>";
        for($j=0;$j<$num3;$j++)
        {
        echo"<td>$row[$j]</td>";    
        }
        echo"</tr>";
        }//for
        echo"</table>";
        $y=$_POST['myid'];
        echo"<input type='hidden' name='negin' value='$y'>";
        }//if


//showing pharmacies($_POST['ph']):
    $query2="select * from pharmacies";
    $result2=mysql_query($query2);
    $nump=Mysql_num_rows($result2);

    echo "Please Select a Pharmacy:<select ID=2 name='ph'>";
    echo"<option >select please";
    for($i=0;$i<$nump;$i++)
    {
    $row=mysql_fetch_row($result2); 
    echo"<option value=$row[1]>$row[1]";
    echo"</option>";
    }
echo"</SELECT>";

//showing drugs($_POST['dg']):
$query2="select * from pharmacy";
$result2=mysql_query($query2);
$nump=Mysql_num_rows($result2);

echo "Please Select Drug:<select ID=1 name='dg'>";
echo"<option >select please";
for($i=0;$i<$nump;$i++)
{
$row=mysql_fetch_row($result2); 
echo"<option  >$row[0]";
echo"</option>";
}
echo"</SELECT>";    

echo"<b>Quantity:<input type='text' name='txt1'/>";
echo"<input type='submit' name='insert' value='insert this drug'/>";

}//ifmyid

if(array_key_exists('insert',$_POST))
{
    echo "HELLO";
    $negin=$_POST['negin'];
    $qname="select * from pnt where id='$negin'";

    $resname=mysql_query($qname);
    $rown=mysql_fetch_row($resname);
    echo $qname;
    $na=$rown[1];
    mysql_real_escape_string ($_POST['dg']);
    mysql_real_escape_string ($_POST['txt1']);
    mysql_real_escape_string ($_POST['ph']);
    mysql_real_escape_string ($na);
    $ins="insert into request(drug,qty,ph,situation,Doctor,userp)values('".$_POST['dg']."',".$_POST['txt1'].",'".$_POST['ph']."','underprocess','$uname','$na')";
    echo $ins;
    $rlt=mysql_query($ins);
    //showing prescribe(table request)
    $in="select * from request";
    $rslt=mysql_query($in);
    $num2=Mysql_num_rows($rslt);
    $num3=Mysql_num_fields($rslt);  
        if($num2>0)
        {
        echo "<table border=2>";
        echo"<td>id</td><td>drug</td><td>quantity</td><td>Doctor</td><td>explanation</td><td>pharmacy</td>";
        for($i=0;$i<$num2;$i++)
        {
        $row=mysql_fetch_row($rslt);
        echo "<tr>";
        for($j=0;$j<$num3;$j++)
        {
        echo"<td>$row[$j]</td>";    
        }
        echo"</tr>";
        }//for
        echo"</table>";
        }//if$num2
}

Answer 1

在同一个表单中使用两个提交按钮

这是你的问题吗？我不知道 。如果它那么简单!

<form name="xx" method="post">
<input type="submit" name="submit1" value="1" />
<input type="submit" name="submit2" value="2" />
</form>

第 1 步:在表单中添加同名的提交按钮。

if($_POST['submit1']=="1")
{
echo "form submited using first submit button";
}

if($_POST['submit2']=="2")
{
echo "form submited using second submit button";
}

第 2 步:在不同的部分获取值。然后你可以为每个按钮做不同的工作

我觉得有帮助!