問題$ _ POST

Question

我得到了我的問題：D是約$ _ POST使用我的ID是不能在插入訪問提交按鈕，我插入隱藏在，如果我的ID

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

的聲明使用兩個提交在相同的形式

按鈕是你的問題？我不確定 。如果它那麼簡單！

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

第1步：在窗體中添加具有相同名稱的提交按鈕。

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：在不同部分中獲取值。然後，你可以爲每個按鍵

我認爲它可以幫助做不同的工作！