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javascript - 尝试从选择选项将数据拉入表中

转载 作者：行者123 更新时间：2023-11-29 23:52:35 26 4
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对于这个网站，我是个新手，但我认为这将是帮助解决我的问题的最佳机会。现在，我正在尝试从所选选项中提取数据并将其放入页面上的表格中。然而，当我完成代码并将其用于测试时，当我点击选择选项时，屏幕只是空白并且没有发布任何数据。代码附在下面。 PHP代码附在下面。

{<?php

$q = strval($_GET['q']);

$con = mysqli_connect('localhost', '$$$$', '', 'plat');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,'platform');
$sql =  "SELECT DISTINCT cert_type,
            status
            FROM em_certification_type 
            WHERE Id = '".$q."'";
$result = mysqli_query($con,$sql);

if($result === FALSE)   {
>die(mysql_error());
}

echo "<table border='1'>";

while($row = mysqli_fetch_array($result))   {
    echo "<tr>";
    echo "<td>" . $row['cert_type'] . "</td>";
    echo "<td>" . $row['status'] . "</td>";
    echo "</tr>";
}

echo "</table>";

?>}

HTML(需要一些清理)

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8">
<?php
/*establishing connection*/
$con = mysql_connect("localhost", "^^^^") or die("Connection Failed");
$db = mysql_select_db("plat")or die("Database not available");


/*setting up query for select box*/

$query1 = "SELECT DISTINCT SpecialtyCode
        FROM em_certification_type
        WHERE SpecialtyCode != ''";

$result1 = mysql_query($query1);


if (!$result1)  {
    $message = 'Invalid query on general requirements: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query1;
    die($message);
}


/*setting query for general requirements table*/
$query2 = "SELECT DISTINCT cert_type,
            expiration_date,
            status,
            SpecialtyCode
            FROM em_certification_type 
            CROSS JOIN em_certification 
            WHERE em_certification_type.certification_id = em_certification.certification_type_id
            ORDER BY em_certification_type.certification_id";
$result2 = mysql_query($query2);

if (!$result2)  {
    $message = 'Invalid query on general requirements: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query2;
    die($message);
}

/* for the state requirements*/
$query3 = "SELECT cert_type,
                expiration_date,
                status,
                state
                FROM em_certification_type
                CROSS JOIN em_certification
                WHERE em_certification_type.certification_id = em_certification.certification_type_id
                AND em_certification.state != ''";
$result3 = mysql_query($query3);



if (!$result3)  {
    $message = 'Invalid query on state requirements: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query3;
    die($message);
}

?>
<h3>Credential</h3>

<form action="Cred.php">
<select name="selectSpecialty" class="selectSpecialty" onchange="showCred(this.value)">
<option value="" selected="selected">Choose Specialty</option>
<?php
/*Setting up a while loop to cycle through for the options*/
while ($row = mysql_fetch_array($result1))  
    echo("<option value = '" . $row['SpecialtyCode'] . "'>" . $row['SpecialtyCode'] . "</option>");
?>
</select>
</form>
<br>
<div id="txtHint"><b>Information will be listed here for general creds.</b></div>

</body>
<script src="http://code.jquery.com/jquery-1.11.0.min.js" type="text/javascript" charset="utf-8"></script>
<script src="buttons.js" type="text/javascript" charset="utf-8"></script>
</html>'

最后，JavaScript

function showCred(str)  {
    if (str=="")    {
        document.getElementById("txtHint").innerHTML="";
        return;
    }

    if (window.XMLHttpRequest)  {
        // code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    }
        else    { //code for IE6, IE 5
            xmlhttp=new ActiveXObject("Microsoft.XMLHTTP")
        }
    xmlhttp.onreadystatechange=function()   {
        if (xmlhttp.readyState==4 && xmlhttp.status==200)   {
            document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
        }
    }
    xmlhttp.open("GET","getcred.php?q="+str,true);
    xmlhttp.send();
    }

非常感谢任何想法或建议。预先非常感谢您。

最佳答案

试试这个

更改这部分

$con = mysqli_connect('localhost', '$$$$', '', 'plat');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,'platform');

至

$con = mysqli_connect('localhost', '$$$$', '', 'platform');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

在mysqli中，您可以在mysqli_connect本身中定义数据库。不需要mysqli_select_db()

关于javascript - 尝试从选择选项将数据拉入表中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25540833/

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