Php 简单MYSQL数据未显示致命错误_Php_Mysql_Database

Php 简单MYSQL数据未显示致命错误

php mysql database

Php 简单MYSQL数据未显示致命错误,php,mysql,database,Php,Mysql,Database,我试图在html数据中显示来自mysql数据的数据，并观看了视频，随后仅更改必要的内容，如表名、数据库名等。我老是犯这个错误分析错误：语法错误，第37行C:\xampp\htdocs\keytracker\keytracker.php中意外的“$sql”（T_变量） <html> <head> <title>SRE | Key Tracker</title> <link rel = "stylesheet" type="text/c

我试图在html数据中显示来自mysql数据的数据，并观看了视频，随后仅更改必要的内容，如表名、数据库名等。我老是犯这个错误

分析错误：语法错误，第37行C:\xampp\htdocs\keytracker\keytracker.php中意外的“$sql”（T_变量）

   <html>
<head>
<title>SRE | Key Tracker</title>
<link rel = "stylesheet" type="text/css" href="style.css">
</head>
<body>
<div class="searchandregister">
<h2 class="cat-title">Search Booked Keys</h2>
<form method="POST" action="">
<input class="field" type="text" placeholder="Key ID" onfocus="this.placeholder=''" onblur="this.placeholder='Key ID'">
<input class="field" type="text" placeholder="Property Address" onfocus="this.placeholder=''" onblur="this.placeholder='Property Address'">
<input class="field" type="text" placeholder="Name" onfocus="this.placeholder=''" onblur="this.placeholder='Name'">
<input class="field" type="text" placeholder="Creditor" onfocus="this.placeholder=''" onblur="this.placeholder='Creditor'">
<br/><input class="button" type="submit" value="Search">
</form><br/>


<h2 class="cat-title">Book Out New Key</h2>
<form method="POST" action="">
<input class="field" type="text" placeholder="Key ID" onfocus="this.placeholder=''" onblur="this.placeholder='Key ID'">
<input class="field" type="text" placeholder="Property Address" onfocus="this.placeholder=''" onblur="this.placeholder='Property Address'">
<input class="field" type="text" placeholder="Name" onfocus="this.placeholder=''" onblur="this.placeholder='Name'">
<input class="field" type="text" placeholder="Creditor" onfocus="this.placeholder=''" onblur="this.placeholder='Creditor'">
<br/><input class="button" type="submit" value="Register">
</form>

</div>

<div class="results">
<h2 class="cat-title-right">Results</h2>
<?php
    //Make connection
        mysql_connect('localhost', 'access', 'AR51Bigwater');
    //Select DB
        mysql_select_db('keytracker')

    $sql="SELECT * FROM keys";
    $records = mysql_query($sql);
    //Display Data

?>
<table cellpadding="1" cellspacing="1">
<tr>

<th>ID</th>
<th>Name</th>
<th>Company</th>
<th>Out_Date</th>
<th>Due_Date</th>
<tr>

<?php
while ($keys=mysql_fetch_assoc($records)){
    echo "<tr>";
    echo "<td>".$keys['ID']."</td>";
    echo "<td>".$keys['Name']."</td>";
    echo "<td>".$keys['Company']."</td>";
    echo "<td>".$keys['Out_Date']."</td>";
    echo "<td>".$keys['Due_Date']."</td>";
    echo "</tr>";
}   
?>
</table>
</div>


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这是根据错误发生的部分：

 <?php
    //Make connection
        mysql_connect('localhost', 'access', 'AR51Bigwater');
    //Select DB
        mysql_select_db('keytracker')

    $sql="SELECT * FROM keys";
    $records = mysql_query($sql);
    //Display Data

?>


由于mysql_select_db（'keytracker'）中缺少分号，因此将上述代码放入您的代码中
 该错误清楚地表明语法错误正如@HoboSapiens所说，您遗漏了第36行中的分号
注意：您应该使用/，而不是mysql。从PHPV5.5.0开始，整个ext/mysql PHP扩展就被正式弃用，将来将被删除
你可以从头开始
前一行缺少分号
<?php
//Make connection
    mysql_connect('localhost', 'access', 'AR51Bigwater');
//Select DB
    mysql_select_db('keytracker');

$sql="SELECT * FROM keys";
$records = mysql_query($sql);
//Display Data

<?php
    //Make connection
    $connection = mysqli_connect('localhost', 'access', 'AR51Bigwater','keytracker');

    $sql="SELECT * FROM keys";
    $records = mysqli_query($connection,$sql);
    //Display Data

?>