【发布时间】：2016-02-04 11:26:00
【问题描述】：
我正在尝试插入 4 个相同的表单。但使用 PHP 与 mysql 具有不同的值。
当我提交数据时，数据库仅从最后一个表单中获取值并插入 4 次。我正在尝试在提交时从所有 4 个中获取值。
<div class="req3">
<h1>Requirement 4</h1>
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<br>
Enter info for 4 teams and it will inserted into the database<br><br>
<div class="sqlForm">
<p class="formHead">Team 1</p>
<label>Team Name:</label> <input type="text" name="teamname"><br>
<label>City:</label> <input type="text" name="city"><br>
<label>Best Player:</label> <input type="text" name="bestplayer"><br>
<label>Year Formed:</label> <input type="text" name="yearformed"><br>
<label>Website:</label> <input type="text" name="website"><br>
</div>
<div class="sqlForm">
<p class="formHead">Team 2</p>
<label>Team Name:</label> <input type="text" name="teamname"><br>
<label>City:</label> <input type="text" name="city"><br>
<label>Best Player:</label> <input type="text" name="bestplayer"><br>
<label>Year Formed:</label> <input type="text" name="yearformed"><br>
<label>Website:</label> <input type="text" name="website"><br>
</div>
<div class="sqlForm">
<p class="formHead">Team 3</p>
<label>Team Name:</label> <input type="text" name="teamname"><br>
<label>City:</label> <input type="text" name="city"><br>
<label>Best Player:</label> <input type="text" name="bestplayer"><br>
<label>Year Formed:</label> <input type="text" name="yearformed"><br>
<label>Website:</label> <input type="text" name="website"><br>
</div>
<div class="sqlForm">
<p class="formHead">Team 4</p>
<label>Team Name:</label> <input type="text" name="teamname"><br>
<label>City:</label> <input type="text" name="city"><br>
<label>Best Player:</label> <input type="text" name="bestplayer"><br>
<label>Year Formed:</label> <input type="text" name="yearformed"><br>
<label>Website:</label> <input type="text" name="website"><br><br></div>
<input class="styled-button" type="submit" name="insert" value="Submit">
</form>
<?php
if (isset($_POST['insert'])) {
insertTable();
} else {
$conn->close();
}
function insertTable() {
$servername = "localhost:3306";
$username = "XXXXX";
$password = "XXXXX";
$dbname = "XXXXX";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
echo ("Connection failed: " . $conn->connect_error);
} else {
$varTname = $_POST['teamname'];
$varCity = $_POST['city'];
$varBplayer = $_POST['bestplayer'];
$varYearformed = $_POST['yearformed'];
$varWebsite = $_POST['website'];
$sql = "INSERT INTO Teams (teamname, city, bestplayer, yearformed, website)
VALUES ('$varTname', '$varCity', '$varBplayer', '$varYearformed', '$varWebsite'),
('$varTname', '$varCity', '$varBplayer', '$varYearformed', '$varWebsite'),
('$varTname', '$varCity', '$varBplayer', '$varYearformed', '$varWebsite'),
('$varTname', '$varCity', '$varBplayer', '$varYearformed', '$varWebsite')";
if ($conn->multi_query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
mysql_query($sql);
function PrepSQL($value)
{
// Stripslashes
if(get_magic_quotes_gpc())
{
$value = stripslashes($value);
}
// Quote
$value = "'" . mysql_real_escape_string($value) . "'";
return($value);
}
}
}
?>
【问题讨论】：
-
用不同的名称命名您的控件，例如 teamname1、teamname2、..
标签： php mysql forms insert sql-insert