doulu7174 2016-01-21 23:16
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这个简单的代码Checkbox从PHP到MYSQL的数组有什么问题?

I am not sure what I am missing with this Array output of php to mysql. It really seems simple and I am missing something. I would think it would break it up into checkbox[1], checkbox[2], checkbox[3].

<html>
<body>
<form method="post" action="output_to_sql.php">
Check box - Please choose type of residence:<br />
Steak:<input type="checkbox" value="Steak" name="checkbox[]"><br />
Pizza:<input type="checkbox" value="Pizza" name="checkbox[]"><br />
Chicken:<input type="checkbox" value="Chicken" name="checkbox[]"><br />
<input type="submit" value="submit" name="submit">
</form>

output_to_sql.php

<?php
$servername = "localhost";
$username = "user";
$password = "password";
$dbname = "database";

// Create connection
$connection = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
echo "success";
}
$checkbox = $_POST["checkbox"];
if (!isset($_POST['submit'])) { 
// if page is not submitted to itself echo the form
echo "error submitting form";
} 
else //show form data with check box answers
{
//make an array with check boxes and show result
foreach ($checkbox as $checkboxitems) {
echo $checkboxitems."<br />";
}
$sql =  mysqli_query("INSERT INTO `table` (`checkbox1`, `checkbox2`, `checkbox3`) VALUES ('$checkboxitems[0]','$checkboxitems[1]','$checkboxitems[2]')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?> 
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1条回答 默认 最新

  • dozc58418381 2016-01-21 23:22
    关注

    Only checked checkboxes are successful controls.

    So if no checkbox is checked, then the array will be empty.

    If one is checked, then it will have one item in it (at index 0 not index 1, arrays start counting from 0).


    In general, this should be resolved with a many-to-many relationship.

    After inserting your data (without checkboxes) into your "table" table, loop over $_POST['checkboxitems'] and insert the id of your new row and the id of the matching foodstuff in the "foodstuffs" table into your linking table.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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