Php 如何将复选框的值插入数据库？
我这里有我的代码，关于我的复选框，但我得到了一些错误，当我点击我的提交按钮。虽然它会打印我在复选框中选择的所有值，但我的sql脚本中有一个错误，警告：mysqli_query至少需要2个参数，1个参数在第21行的C:\xampp\htdocs\project\candidate\president2.php中给出。我只想保存我在数据库中选择的值。请帮忙Php 如何将复选框的值插入数据库？,php,mysqli,Php,Mysqli,我这里有我的代码，关于我的复选框，但我得到了一些错误，当我点击我的提交按钮。虽然它会打印我在复选框中选择的所有值，但我的sql脚本中有一个错误，警告：mysqli_query至少需要2个参数，1个参数在第21行的C:\xampp\htdocs\project\candidate\president2.php中给出。我只想保存我在数据库中选择的值。请帮忙 <?php session_start(); ?> <?php //se
<?php session_start(); ?>
<?php
//server info
$server = 'localhost';
$user = 'root';
$pass = 'root';
$db = 'user';
// connect to the database
$mysqli = new mysqli($server, $user, $pass, $db);
// show errors (remove this line if on a live site)
mysqli_report(MYSQLI_REPORT_ERROR);
?>
<?php
if ($_POST['representatives']){
$check = $_POST['representatives'];
foreach ($check as $ch){
//this is my line 21 error. what i want here is to save the selected checkbox into my database but i got some error and i couldnt save it to my database
mysqli_query("INSERT INTO sample (name) VALUES ('". $ch ."') ");
echo $ch. "<br>";
}
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<html>
<head>
<script type="text/javascript">
<!--
function get_representatives_value()
{
for (var i=0; i < document.list.representatives.length; i++)
{
if (document.list.representatives[i].value = true)
{
return document.getElementById('txt').innerHTML =document.list.representatives[i].value
}
}
}
//-->
</script>
title></title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
<link href="candidate.css" rel="stylesheet" type="text/css">
</head>
<body> <p id="txt"></p>
<form name="list" action="president2.php" method="post" onSubmit="return get_representatives_value()">
<div id="form">
<?php
// get the records from the database
if ($result = $mysqli->query("SELECT * FROM candidate_info WHERE position= 'representatives' AND department ='CCEITE' ORDER BY cand_id"))
{
// display records if there are records to display
if ($result->num_rows > 0)
{
// display records in a table
echo "<table border='1' cellpadding='10'>";
// set table headers
echo "<tr><th>Student ID</th><th>Candidate ID</td><th>Course</th><th colspan = '3'>Name</th></tr>";
while ($row = $result->fetch_object())
{
// set up a row for each record
echo "<tr>";
echo "<td>" . $row->cand_studid . "</td>";
echo "<td>".$row->cand_id."</td>";
echo "<td>" . $row->course . "</td>";
echo "<td coslpan ='5'>" . $row->fname . " ". $row->mname ." ". $row->lname ." </td>";
echo "<td><input type ='checkbox' name='representatives[]' id='". $row->studid ."' value='" . $row->fname . " ". $row->mname ." ". $row->lname . "'onchange='get_representatives_value()' /></td>";
echo "</tr>";
}
echo "</table>";
}
// if there are no records in the database, display an alert message
else
{
echo "No results to display!";
}
}
// show an error if there is an issue with the database query
else
{
echo "Error: " . $mysqli->error;
}
// close database connection
$mysqli->close();
echo "<input type='submit' name='representatives value='Submit' />";
?>
</div>
</form>
</body>
</html>
mysqli_查询函数要求$mysqli链接作为第一个参数。有两种方法可以修复错误。下面是错误 要解决这个问题，只需将其更改为下面两个>Id中的一个，使用第一个选项，因为您已经在代码中的某个地方使用了它
$mysqli->query("INSERT INTO sample (name) VALUES ('". $ch ."') ");
看看我的例子，我希望它能帮助你-
<?php
if(isset($_POST['team']))
{
foreach($_POST['team'] as $value){
$insert=mysql_query("INSERT INTO team('team') VALUES ('$value')");
}
}
?>
<html>
<body>
<form method="post" action="lol.php">
<input type="checkbox" name="team[]" value="IN"> India<br />
<input type="checkbox" name="team[]" value="DK"> Dark <br />
<input type="checkbox" name="team[]" value="LA"> lolax <br />
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
后跟作为第二个参数的查询。您似乎没有在代码中使用这两个函数。看到您声明了一个mysqli对象，您可能打算改为使用$mysqli->query。看起来您混合了OOP和过程mysqli函数。mysqli\u查询的第一个参数应该是link或use$link->see此doc@njk&farmer1992。。谢谢你的建议，现在我得到了我需要的。。。thaks很多..问题出在mysqli身上。虽然这确实解释了如何使用不推荐使用的mysql_uu函数，但它并没有解决这里的问题。@TJohnW。现在我的问题是如何计算被选中的候选人。如果您能建议我怎么做，我将不胜荣幸，先生。。。但再次感谢你的帮助。。。
mysqli_query($mysqli, "INSERT INTO sample (name) VALUES ('". $ch ."') ");
<?php
if(isset($_POST['team']))
{
foreach($_POST['team'] as $value){
$insert=mysql_query("INSERT INTO team('team') VALUES ('$value')");
}
}
?>
<html>
<body>
<form method="post" action="lol.php">
<input type="checkbox" name="team[]" value="IN"> India<br />
<input type="checkbox" name="team[]" value="DK"> Dark <br />
<input type="checkbox" name="team[]" value="LA"> lolax <br />
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>