Question

我希望我的第一个字段（entrynum）在更新时执行MySQLi搜索，如果发现记录自动填充其余字段。

我已更新问题和原帖。我有一个表单字段，onchange正确触发，但无论我尝试什么都不返回任何数据。

reg.php

<?php include("process.php"); ?>

<!doctype html>
<html>
<head>
    <script type="text/javascript" src="js/jquery.js"></script>
    <script type="text/javascript" src="js/entrynum.js"></script>
</head>
<body>
        <?php 
if (isset($_POST['reg-submit'])) {
    echo "<p id='notice' style='padding: .5em; border: 2px solid red;'>Entry $entrynum Saved!<br>$timenow on $datenow<br><a href='upload.php' style='font-size:xx-large;'>Upload Pictures</a></p>";
} else {
    echo "<p id='notice' style='display: none; padding: .5em; border: 2px solid red;'></p>";
}
?>
    <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
        <fieldset>
            <legend><h1>Registration</h1></legend>
            <label for="entrynum">Entry Number</label>
            <input type="number" pattern="\d*" name="entrynum" id="entrynum" value="" required="true" placeholder="" autofocus onchange="entry_check()" />

            <label for="fname">First Name</label>
            <input type="text" name="fname" id="fname" value="" required="true" placeholder="" list="" style="text-transform:capitalize" onkeyup="javascript:this.value=this.value.charAt(0).toUpperCase() + this.value.substr(1);" />

            <label for="lname">Last Name</label>
            <input type="text" name="lname" id="lname" value="" required="true" placeholder="" list="" style="text-transform:capitalize" onkeyup="javascript:this.value=this.value.charAt(0).toUpperCase() + this.value.substr(1);" />

            <input type="submit" name="reg-submit" id="reg-submit" value="Submit" />
        </fieldset> 
    </form>
</body>
</html>

process.php

<?php
include("connect.php");
error_reporting(E_ALL);
ini_set('display_errors', 'On');
date_default_timezone_set('US/Central');
session_start();

$datenow = date("y-m-d");
$timenow = date("h:i:sa");

// registration
if (!empty($_POST['reg-submit'])) {
    $entrynum = $_POST['entrynum'];

    $_SESSION['entrynum'] = $entrynum;

    $fname = $_POST['fname'];
    $lname = $_POST['lname'];


    $sql = "INSERT INTO HeatWaveData (entrynum, FName, LName)
        VALUES ('$_POST[entrynum]','$_POST[fname]','$_POST[lname]')";

    if (!$db->query($sql)) { die("Error: {$db->errno} : {$db->error}"); }

}
?>

entrynum.js

function entry_check() {


    var entrynum = $("#entrynum").val();

    // Send the AJAX call
    $.post(
        'entrysearch.php', // TO search.php page
        {entrynum: entrynum}, // the DATA sent with the request
        function(data) { // a CALLBACK function
            if (data == 'none') { // no rows were found or an error occurred
                document.getElementById("notice").innerHTML = "New Entry!";
                document.getElementById("notice").style.display = "block";
                return;
            } else {
                document.getElementById("notice").innerHTML = "Already Exists!";
                document.getElementById("notice").style.display = "block";
            }
            data = JSON.parse(data); // parse the array returned from the server

            // set the data in the matching fields (could be done manually if needed)
            // for (var i = 0; i < data.length; i++) {
            //  $("#data" + i).val(data[i]);
            // }
        }
    );

}

entrysearch.php

<?php
// check post data is received
if (!isset($_POST['entrynum'])) {
    echo 'none';
    die();
}
// create a prepared statement against sql-injection
$sql = $db->prepare("SELECT * FROM HeatWaveData WHERE entrynum=%d", $_POST['entrynum']);
// execute the query (depending on your db class type get the results otherwise)
$results = $db->query($sql);
// After this line results should be a matrix (change/add lined as needed)
// we need the first row returned (probably only 1 was returned...)
$result = $results[0];
// check if a row was returned
if (!$result) {
    echo 'none';
    die();
}
echo json_encode($result);
?>

Answer 1

您需要做的是输入字段的onChange函数，该函数将向脚本发送ajax调用，该脚本将搜索您需要的数据。
然后使用JavaScript，您可以更新相关字段。

我正在这里写一个代码示例（没有选中），但是玩它并且它应该工作：
注意：我在我的代码中使用jQuery

foo.html

    <script>
    function input_change() {
        // Get the value from the input
        var entrynum = $("#entrynum").val();

        // Send the AJAX call
        $.post(
                'search.php', // TO search.php page
                {entrynum: entrynum}, // the DATA sent with the request
                function(data) { // a CALLBACK function
                    if (data == 'none') { // no rows were found or an error occurred
                        return;
                    }
                    data = JSON.parse(data); // parse the array returned from the server

                    // set the data in the matching fields (could be done manually if needed)
                    for (var i = 0; i < data.length; i++) {
                        $("#data" + i).val(data[i]);
                    }
                }
        );
    }
</script>

<input type="number" pattern="\d*" name="entrynum" id="entrynum" value="" required="true" placeholder="" autofocus onchange="input_change()" />

<input type="text" name="data0" id="data0">
<input type="text" name="data1" id="data1">
<input type="text" name="data2" id="data2">
<input type="text" name="data3" id="data3">

的search.php

<?php
//------------(maybe different depending on your db class type)----------------
// begin sesstion and connect to db and stuff
$db = //get database connection
// check post data is received
if (!isset($_POST['entrynum'])) {
    echo 'none';
    die();
}
// create a prepared statement against sql-injection
$sql = $db->prepare("SELECT * FROM table WHERE entrynum=%d", $_POST['entrynum']);
// execute the query (depending on your db class type get the results otherwise)
$results = $db->query($sql);
// After this line results should be a matrix (change/add lined as needed)
// we need the first row returned (probably only 1 was returned...)
$result = $results[0];
// check if a row was returned
if (!$result) {
    echo 'none';
    die();
}
echo json_encode($result);
?>

无论我尝试什么，sql都不返回数据

1 个答案: