如何将html输入标签的值传递给php代码？

Question

我的patient_display.php代码调用了传递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，抛出下面的错误 SCREAM：忽略 注意屏蔽错误：未定义指数：P_ID在

Answer 1

使用$ _GET和$彦博它的值设置是否（存在）之前只是检查或不。 您可以使用'@'，'isset'，'strlen'和'！empty'。 所以您的病情会

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

和同为$_GET['p_id']

Answer 2

我想你的SQL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> 

而且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';");