dongyongmin5711 2018-02-24 10:51 采纳率: 100%
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PHP：在OOP上使用POST并存储在MySql中

Ok im kinda new to using oop on php, So this what im tryna do. getting the value of the textbox putting them in an array then storing them in database but the problem is it doesnt store the data

<?php
require ('employmentfunc.php');
?>
<html>
<head>
</head>
<body>
<?php 
$arr = array(isset ($_POST['company']), isset ($_POST['position']));

if (isset($_POST['ok']))
{
    $a = new empfunc();
    $a->SaveA($arr);
}
?>
<form name="emp" action="employmentform.php" method="POST">

<TD><input type="text" name="company">
<td><input type="text" name="position">
<input type="submit" class="button1" name="ok">
</form>
</body>
</html>



<?php
class empfunc{

function saveA($array){
    $sqlconnect = mysqli_connect("localhost", "root","","employment");
    $sqlquery = "INSERT INTO info(`Company`, `Position`) VALUES ('$array[0]', '$array[1]')";
    mysqli_query ($sqlconnect, $sqlquery);
}
}
?>
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  • dro44817 2018-02-24 11:14
    关注

    Your problem is the way you are building the $arr variable. You are storing boolean results from the isset function calls instead of storing the data itself. I would rewrite that part of the code using a few more lines to make it more readable. Since you are storing the result of isset your values stored in the database will be either 0 or 1. Instead, either do not use the isset function or else change them to a ternary operator to store some default data if they are not set:

    $arr = array(
    isset ($_POST['company']) ? $_POST['company'] : 'No Company', 
    isset ($_POST['position']) ? $_POST['position'] : 'No Position'
);

    or if you just want to skip the isset validation:

    $arr = array($_POST['company'], $_POST['position']);
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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