使用HTML / PHP表单访问数据库并输出到同一页面

Question

I have a database of schools, I would like to make it so that when you search the database for existing data, it outputs right next the the input box. So without going to a new page. Here's what a have:

<form action=" " method="post">
School's name: <input type="text" name="schoolname"> <br/>
<input type="submit" name="button" value="Search">
</form>

<?php
$school      = $_POST['schoolname'];

$conn = mysql_connect("localhost", "root");
mysql_select_db("finalproject");

$sql = "select * from presentations where school like '%$school%'";

$result = mysql_query($sql, $conn) or die(mysql_error());

if ( mysql_num_rows($result) >0)
    {
    while ($newArray = mysql_fetch_array($result))
        {
        $school  = $newArray['school'];
        $date = $newArray['date'];
        $place  = $newArray['place'];
        $time = $newArray['time'];


        echo $school . ", " . $place . ", " . $date . ", " . $time . "<br />" ;
        }
    }
    else 
        {
        echo "Record not found";
        }



mysql_close($conn);
?>

This is code that I have used previously to link to another page, outputting there. but now I just want to output it on the same page. I did move some code over from the other page which no longer seems to be working. The PHP bit just outputs: "0) { while ($newArray = mysql_fetch_array($result)) { $school = $newArray['school']; $date = $newArray['date']; $place = $newArray['place']; $time = $newArray['time']; echo $school . ", " . $place . ", " . $date . ", " . $time . " " ; } } else { echo "Record not found"; } mysql_close($conn); ?>" onto my page below the input. I'm really new to this, so anyones help would be greatly appreciated. :D

Answer 1

您必须使用Ajax向服务器发出请求，而无需重新加载页面。

Answer 2

You could modify this little snippet for the ajax part...It's using jQuery, so you'll need to include the library into your page.

$('#submitButtonID').click(function(){

var data = {
    schoolName: $('#schoolName').val()
};
$.ajax({
    url: "PhpPageWithQuery.php",
    type: "post",
    data: data,
    success: function(msg) {
        $('#resultsDiv).html(msg);

    }
});

Answer 3

Make sure that your file extension is .php . Also check if $_POST['schooname'] isset and if it is then continue with php code. A few other pointers, the use of the mysql extension is not recommended since it's deprecated. Use either mysqli or PDO , also santize your input if you must use it. I left the sanitation bit to you.

<form action=" " method="post">
School's name: <input type="text" name="schoolname"> <br/>
<input type="submit" name="button" value="Search">
</form>

<?php

if( isset( $_POST['schoolname'] ) && strlen( trim( $_POST['schoolname'] ) ) > 0  )
{
        $school      = $_POST['schoolname'];

        $conn = mysql_connect("localhost", "root");
        mysql_select_db("finalproject");

        $sql = "select * from presentations where school like '%$school%'";

        $result = mysql_query($sql, $conn) or die(mysql_error());

        if ( mysql_num_rows($result) >0)
        {
            while ($newArray = mysql_fetch_array($result))
            {
                $school  = $newArray['school'];
                $date = $newArray['date'];
                $place  = $newArray['place'];
                $time = $newArray['time'];


                echo $school . ", " . $place . ", " . $date . ", " . $time . "<br />" ;
            }
        }
        else 
        {
            echo "Record not found";
        }



        mysql_close($conn);
}
?>