Question

我的patient_display.php代码调用传递P_ID的patient_history.php文件。以下是我的代码。

patient_display.php:
echo '<form name="Patient" action="patient_history_display.php" method="get">';
$pid=$_GET["patient_id"];
echo '<input type="text" name="p_id" value= '.$pid.' >';
</form>

patient_history.php:
$result = mysqli_query($con,"SELECT P.P_F_NAME, P.P_L_NAME,P.P_ADDR,     round(datediff(now(),P.P_DOB)/365) AS P_AGE, D.D_DESC, A.A_DESC 
FROM P_HAS_A  PA, patient P, P_HAS_D PD, n_provide_m NM, disease D, allergy A 
WHERE P.P_ID = PD.P_ID AND PD.D_ID = D.D_ID AND P.P_ID = PA.P_ID AND PA.A_ID = A.A_ID AND P.P_ID='{$_GET["p_id"]}';");
$pid=$_GET["p_id"];

然而，它抛出以下错误 SCREAM：忽略错误抑制注意：未定义的索引：

中的p_id

Answer 1

在使用$ _GET和$ _POST之前检查它的值是否设置（存在）。你可以使用'@'，'isset'，'strlen'和'！empty'。所以你的情况将是

 if(isset($_GET['patient_id'])) { 
   $id = $_GET['patient_id'];
   echo '<input type="text" name="p_id" value= '.$pid.' >';
 }

和$_GET['p_id']

相同

Answer 2

我认为你在sql中有错误。试试这个：

$pid=$_GET["p_id"];
$result = mysqli_query($con,"SELECT P.P_F_NAME, P.P_L_NAME,P.P_ADDR,     round(datediff(now(),P.P_DOB)/365) AS P_AGE, D.D_DESC, A.A_DESC 
FROM P_HAS_A  PA, patient P, P_HAS_D PD, n_provide_m NM, disease D, allergy A 
WHERE P.P_ID = PD.P_ID AND PD.D_ID = D.D_ID AND P.P_ID = PA.P_ID AND PA.A_ID = A.A_ID AND P.P_ID='$pid'");

Answer 3

试试这个

echo '<form name="Patient" action="patient_history.php" method="get">';
$pid=$_GET["p_id"];
echo '<input type="text" name="p_id" value= '.@$pid.' >';
</form>

和patient_history.php：

$pid = $_GET['p_id'];
$result = mysqli_query($con,"SELECT P.P_F_NAME, P.P_L_NAME,P.P_ADDR,
round(datediff(now(),P.P_DOB)/365) AS P_AGE, D.D_DESC, A.A_DESC 
FROM P_HAS_A  PA, patient P, P_HAS_D PD, n_provide_m NM, disease D, allergy A 
WHERE P.P_ID = PD.P_ID AND PD.D_ID = D.D_ID AND P.P_ID = PA.P_ID AND PA.A_ID = A.A_ID AND
P.P_ID='{$pid';");

如何将值从html输入标记传递给php代码？

3 个答案: