我的固定代码归功于Marc B

Question

我在一个页面上有一个链接到PHP文件（动作）的表单，现在PHP结果显示在这个PHP文件/页面中。但我希望结果显示在带有表单的页面上。我已经彻底搜索过，无法在任何地方找到它。也许你们中的任何人都可以提供帮助吗？

代码：/citizens.php（主页）

<form method="post" action="/infoct.php">
<input type="text" name="ID" placeholder="ID">
<input name="set" type="submit">
</form>

代码：/infoct.php

<!DOCTYPE html>
<html>
<head>
<!-- <meta http-equiv="refresh" content="0; url=/citizens.php" /> -->
</head>

<body>

<?php {
$ID2 = isset($_POST['ID']) ? $_POST['ID'] : false;
}

$connect = mysql_connect('localhost', 'root', 'passwd'); 
mysql_select_db ('inhabitants'); 
$sql = "SELECT `Name`, `Surname`, `DOB`, `RPS`, `Address` FROM `citizens` WHERE ID = $ID2";
$res = mysql_query($sql);
echo "<P1><b>Citizen Identification number is</b> $ID2 </p1>";
while($row = mysql_fetch_array($res))
{
    echo "<br><p1><b>First Name:  </b></b>", $row['Name'], "</p1>";
    echo "<br><p1><b>Surname:  </b></b></b>", $row['Surname'], "</p1>";
    echo "<br><p1><b>Date of birth:  </b></b></b></b>", $row['DOB'], "</p1>";
    echo "<br><p1><b>Address:  </b></b></b></b></b>", $row['Address'], "</p1>";
    echo "<br><p1><b>Background information:  </b><br>", $row['RPS'], "</p1>";
}
mysql_close ($connect);

?>
</body>
</html>

我的固定代码归功于Marc B

<form method="post">
<input type="text" name="ID" placeholder="ID">
<input name="set" type="submit">
</form>
<?php

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$ID = isset($_POST['ID']) ? $_POST['ID'] : false;

$connect = mysql_connect('fdb13.biz.nf:3306', '1858208_inhabit', '12345demien12345'); 
mysql_select_db ('1858208_inhabit'); 
$sql = "SELECT `Name`, `Surname`, `DOB`, `RPS`, `Address` FROM `citizens` WHERE ID = $ID";
$res = mysql_query($sql);
if ($ID > 0) {
    echo "<p><b>Citizen Identification number is</b>  </p>";

    while($row = mysql_fetch_array($res))
    echo "<br><p><b>Surname:  </b></b></b>", $row['Surname'], "</p>";
    echo "<br><p><b>First Name:  </b></b>", $row['Name'], "</p>";
    echo "<br><p><b>Date of birth:  </b></b></b></b>", $row['DOB'], "</p>";
    echo "<br><p><b>Address:  </b></b></b></b></b>", $row['Address'], "</p>";
    echo "<br><p><b>Background information:  </b><br>", $row['RPS'], "</p>";

mysql_close ($connect);
}
    else {
      echo "<p>Enter a citizen ID above</p>";
    }
}
?>

Database snap DB Snap

Answer 1

单页表单+提交处理程序非常基本：

<?php

if ($_SERVER['REQUEST_METHOD'] == 'POST') { 
 ... form was submitted, process it ...
 ... display results ...
 ... whatever else ...
}
?>

<html>
<body>
<form method="post"> ... </form>
</body>
</html>

真的就是这样。

Answer 2

在同一页面上使用代码（citizens.php）

<?php

if (isset($_POST)) { 
Do manipulation
}
?>

否则使用 ajax 并从表单中删除操作方法。

<form method="post" id="contactForm">
<input type="text" name="ID" placeholder="ID">
<input name="set" type="buttom" id="submitId">
</form>

<script>
$("#submitId").click(function(){
   var Serialized =  $("#contactForm").serialize();
    $.ajax({
       type: "POST",
        url: "infoct.php",
        data: Serialized,
        success: function(data) {
            //var obj = jQuery.parseJSON(data); if the dataType is not specified as json uncomment this
            // do what ever you want with the server response
        },
   error: function(){
        alert('error handing here');
      }
    });
});
</script>

最后在你的infact.php 中回应数据，以便ajax可以获得数据。

Answer 3

您可以将所有内容放在infoct.php中，如下所示：

<!DOCTYPE html>
<html>
<head>
<!-- <meta http-equiv="refresh" content="0; url=/infoct.php" /> -->
</head>

<body>
<form method="post" action="/infoct.php">
<input type="text" name="ID" placeholder="ID" value="<?php isset($_POST['ID']) ? $_POST['ID'] : '' ?>">
<input name="set" type="submit">
</form>
<?php 
    if (isset($_POST['ID'])) {
        $ID2 = $_POST['ID']; // DO NOT FORGET ABOUT STRING SANITIZATION
        $connect = mysql_connect('localhost', 'root', 'usbw'); 
        mysql_select_db ('inhabitants'); 
        $sql = "SELECT `Name`, `Surname`, `DOB`, `RPS`, `Address` FROM `citizens` WHERE ID = $ID2";
        $res = mysql_query($sql);
        echo "<P1><b>Citizen Identification number is</b> $ID2 </p1>";
        while($row = mysql_fetch_array($res))
        {
            echo "<br><p1><b>First Name:  </b></b>", $row['Name'], "</p1>";
            echo "<br><p1><b>Surname:  </b></b></b>", $row['Surname'], "</p1>";
            echo "<br><p1><b>Date of birth:  </b></b></b></b>", $row['DOB'], "</p1>";
            echo "<br><p1><b>Address:  </b></b></b></b></b>", $row['Address'], "</p1>";
            echo "<br><p1><b>Background information:  </b><br>", $row['RPS'], "</p1>";
        }
        mysql_close ($connect);
    }
?>
</body>
</html>

不要忘记字符串清理！

Answer 4

我找到了解决以下问题的方法：

在同一页面上显示结果

感谢Marc B

单页表单+提交处理程序非常基本：

<?php

if ($_SERVER['REQUEST_METHOD'] == 'POST') { 
 ... form was submitted, process it ...
 ... display results ...
 ... whatever else ...
}
?>

<html>
<body>
<form method="post"> ... </form>
</body>
</html>

真的就是这样。

仅显示第一个值

我将此问题添加到我的代码中解决了这个问题：

while($row = mysql_fetch_array($res)) {
$surname=$row['Surname'];
$name=$row['Name'];
$dob=$row['DOB'];
$address=$row['Address'];
$RPS=$row['RPS'];

现在显示的是所有值，而不是仅显示第一个值。

Answer 5

在同一页面上显示结果

我偶然发现了同样的问题我发现您可以只需要另一个文件。 include_once("PATH_TO_FILE")'。

在/citizens.php

<?php include_once="infoct.php" ?>
<form> ... </form> 
<div>
 <?php $yourdata ?>
</div>

$yourdata 应该是html。

不要忘记字符串清理！

确保从表单中删除操作

比在一个文件中包含所有逻辑和HTML更好。

表单和PHP结果显示在同一页面上

我的固定代码归功于Marc B

5 个答案: