將MYSQL數據從PHP傳遞到HTML

Question

我一直在嘗試使用PHP使用來自MYSQL數據庫的數據填充HTML表。 我遇到的問題是HTML文件中的占位符值從不顯示數據形式MYSQL。

HTML文件

<div class="container" style="background-image: url('http://www.MyWebSite.com/OurWorld/EditorBG.png');
-webkit-background-size: cover; -moz-background-size: cover; -o-background-size: cover;
background-size: repeat; padding-bottom: 5px;">
<br />
<div class="mainbody" style="background-image: url('http://www.MyWebSite.com/Test/Notepad.png');
    background-repeat: no-repeat; margin-left: auto; margin-right: auto; margin-bottom: 10px;
    width: 646px; height: 800px !important; min-height: 100%; overflow: hidden;">
    <div class="main-form" style="margin-top: 280px; margin-left: 15px;">
        <form method="GET" action="http://www.MyWebSite./Test/SuggestionSchemeForm.php">
            <table align="center" bgcolor="white" BORDER=2 BORDERCOLOR=Black summary="Submitted Suggestions">
                <caption bgcolor="white" BORDER=2 BORDERCOLOR=Black >Submitted Suggestions</caption>
                    <thead>
                        <tr><th>Name</th><th>Site</th><th>Status</th><th>Date</th></tr>
                    </thead>
                    <tbody>
                        <tr><td><input placeholder="name"></td>   <td><input placeholder="site"></td>   <td><input placeholder="status"></td>   <td><input placeholder="date"></td></tr>
                        <tr><td><input placeholder="name"></td>   <td><input placeholder="site"></td>   <td><input placeholder="status"></td>   <td><input placeholder="date"></td></tr>
                        <tr><td><input placeholder="name"></td>   <td><input placeholder="site"></td>   <td><input placeholder="status"></td>   <td><input placeholder="date"></td></tr>
                        <tr><td><input placeholder="name"></td>   <td><input placeholder="site"></td>   <td><input placeholder="status"></td>   <td><input placeholder="date"></td></tr>            
                    </tbody>
            </table>
        </form>
    </div>
</div>

PHP文件

<?php
Global $USER;
$servername = "";
$username = "";
$password = "";
$dbname = "";
$firstname = $USER->firstname;
$lastname = $USER->lastname;

$testname = "BLEGH";

echo $testname;


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) 
{
   die("Connection failed: " . $conn->connect_error);
}

$sql = mysql_query("SELECT name, site, status, date FROM enquiries")
$result = $conn->query($sql);


while($row = mysql_fetch_array($result MYSQL_ASSOC))
{
   echo "name {$row['name']}".
      "site {$row['site']}".
      "status {$row['status']}".
      "date {$row[date]}";  
}




if ($conn->query($sql) === TRUE) 
{
     echo "New record created successfully";
} 
else 
{
     echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

Header('Location: http://www.MyWebSite.com/index.php');
exit;
?> 

Answer 1

好吧，他們什么也不能顯示。 首先，您要在PHP代碼的末尾進行重定向（使用標頭“ Location”），因此不會將任何變量傳遞給HTML，並且視圖將立即刷新。 其次，HTML本身不存在，您應該在HTML代碼中回顯從mysql獲取的變量。

<?php 
    // stuff..
    $resultArray = [];

    while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
        $resultArray[] = $row;
    }
?>

<!-- (...) -->

<?php foreach ($resultArray as $row) { ?>
<tr>
    <td><input type="text" placeholder="name" name="name" value="<?php echo $row['name']?>" /></td>
    <td><input type="text" placeholder="site" name="site" value="<?php echo $row['site']?>" /></td>
    <td><input type="text" placeholder="status" name="status" value="<?php echo $row['status']?>" /></td>
    <td><input type="text" placeholder="date" name="date" value="<?php echo $row['date']?>" /></td>
</tr>
<?php } ?>

Answer 2

您試圖顯示內容SELECT * FROM但您需要將其插入到mysql中。因此，像這樣嘗試命令INSERT INTO table_name (value1, value2) VALUES ($value1, $value1) ！

Answer 3

您只需要使用自己的連接數據。

<?php

//connection data
$username = "";
$password = "";
$servername = "";
$dbname = "";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

//select query Use ` atleast around date becasue date is a Reserved word for mysql
if ($result = $conn->query("SELECT `name`, `site`, `status`, `date` FROM enquiries")) {

    while($row = $result->fetch_array(MYSQLI_ASSOC))
    {
        $suggestionsArray[] = $row;
    }

    /* free result set */
    $result->close();
}
?> 


<div class="container" style="background-image: url('http://www.MyWebSite.com/OurWorld/EditorBG.png');
-webkit-background-size: cover; -moz-background-size: cover; -o-background-size: cover;
background-size: repeat; padding-bottom: 5px;">
<br />
<div class="mainbody" style="background-image: url('http://www.MyWebSite.com/Test/Notepad.png');
    background-repeat: no-repeat; margin-left: auto; margin-right: auto; margin-bottom: 10px;
    width: 646px; height: 800px !important; min-height: 100%; overflow: hidden;">
    <div class="main-form" style="margin-top: 280px; margin-left: 15px;">
        <form method="GET" action="http://www.MyWebSite./Test/SuggestionSchemeForm.php">
            <table align="center" bgcolor="white" BORDER=2 BORDERCOLOR=Black summary="Submitted Suggestions">
                <caption bgcolor="white" BORDER=2 BORDERCOLOR=Black >Submitted Suggestions</caption>
                    <thead>
                        <tr><th>Name</th><th>Site</th><th>Status</th><th>Date</th></tr>
                    </thead>
                    <tbody>

                    <?  foreach ($suggestionsArray as $row) { ?>
                                <tr><td><? echo $row['name']; ?></td>   <td><? echo $row['site']; ?></td>   <td><? echo $row['status']; ?></td>   <td><? echo $row['date']; ?></td></tr>

                    <?  } ?>



                    </tbody>
            </table>
        </form>
    </div>
</div>

<?  $conn->close();  ?>