我如何在while标记中提交php中的所有字段
[英]How do I submit all fields within a php while tag
问题描述
Im trying to post the generated form to a second file for further processing, however only the last entry is sent to the second file. 
我试图将生成的表单发布到第二个文件进行进一步处理，但是只有最后一个条目发送到第二个文件。 How do I send all information in the loop to the second page? 如何将循环中的所有信息发送到第二页？ Additionally the $_POST['STAMP'] entry is not being posted. 此外，不会发布$_POST['STAMP']条目。 Do I need a different approach to retrieve the DB data? 我是否需要其他方法来检索数据库数据？
<!DOCTYPE html>
<html>
<body>
<form action="payc.php" method="post">
Pay Range
<input type="date" name="start">
<input type="date" name="stop">
<input type="submit">
</form>
</body>
</html>
<?php
$servername = "localhost";
$username = "user";
$password = "Pass";
$dbname = "DB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM TIME WHERE STAMP BETWEEN '". $_POST['start']."' AND '".$_POST['stop']."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<form action=\"test.php\" method=\"post\">";
while($row = $result->fetch_assoc()){
$duration=(($row['T_OUT']-$row['T_IN'])/60)/60;
echo "<fieldset><input type=\"hidden\" name=\"date\" value=\"".$_POST['STAMP']."\"><input name=\"EMP\" type=\"text\" value=\"". $row['EMP']. "\"><input name=\"time\" type=\"text\" value=\"" . $duration. "\"><input name=\"ITEM\" type=\"text\" value=\"" . $row['ITEM']. "\"><input name=\"NOTE\" type=\"text\" value=\"" . $row['NOTE']. "\"></fieldset>";
}
echo "<input type=\"submit\">";
echo "</form>";
} else {
echo "0 results";
}
foreach($_POST as $results){
var_dump($results);
}
$conn->close();
?>
2 个解决方案
解决方案1
0 已采纳 2015-08-31 22:36:26
The problem is that all the fields have the same name at every loop 问题在于，每个字段在每个循环中都具有相同的名称
The solution consists of two parts: 该解决方案包括两部分：
<input name="date[]">
<input name="EMP[]">
<input name="time[]">
<input name="ITEM[]">
<input name="NOTE[]">
Which will identify it as array not one input 哪个将其标识为数组而不是一个输入
Second part at the test.php file Instead of getting one input with $_POST['date'] It will return an array so you have to loop over them test.php文件的第二部分，而不是使用$ _POST ['date']获得一个输入，它将返回一个数组，因此您必须遍历它们
$info = array();
for ( $index = 0 ; $index < count($_POST['date']) ; $index++ )
{
$record = array();
$record['date'] = $_POST['date'][$index];
$record['EMP'] = $_POST['EMP'][$index];
$record['time'] = $_POST['time'][$index];
$record['ITEM'] = $_POST['ITEM'][$index];
$record['NOTE'] = $_POST['NOTE'][$index];
$info[] = $record
}
解决方案2
-2 2015-08-31 19:47:20
1) You are vulnerable to sql injection attacks 1）您容易受到sql注入攻击
2) You never check if the form was submitted at all - your php code runs regardless of HOW the page was loaded. 2）您永远不会检查表单是否已提交-无论页面如何加载，您的php代码都会运行。 At minimum, you should have something like 至少，您应该有类似
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
... db stuf ...
}
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