Question

我正在尝试在html数据中显示来自mysql数据的数据并观看视频并跟踪它们仅更改表名DB名称等所需的内容。我一直收到这个错误？

解析错误：语法错误，第37行的C：\ xampp \ htdocs \ keytracker \ keytracker.php中的意外'$ sql'（T_VARIABLE）

   <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>

这是根据错误发生错误的部分：

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

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

?>

Answer 1

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

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

＆GT;

上面的代码放在您的代码中，因为syntex错误是mysql_select_db（'keytracker'）中缺少分号;

Answer 2

该错误明确指出语法错误您错过了第36行中的分号，如@HoboSapiens所述。

注意：您应该使用MySQLi / PDO_MySQL而不是mysql。自PHP v5.5.0起，整个ext / mysql PHP扩展已正式弃用，将来将被删除。

你可以从here开始。

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

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

?>

简单的MYSQL数据不显示致命错误

2 个答案: