Question

到目前为止，我的代码似乎与一行有所不同，我得到以下错误：

警告：mysql_fetch_assoc（）期望参数1为资源，第37行的C：\ xampp \ htdocs \ Website \ Search.php中给出布尔值

第37行如下：

while($name=mysql_fetch_assoc($records))

其余代码：

HEAD> <TITLE> Search </TITLE> </HEAD>
<BODY>

<h1>Search for your favourite movies here</h1>
<b>`enter code here`
<p> Here you can search through our exciting selection of existing and upcoming movies </p>
<p> You can search our website by choosing from one of the drop down lists below depending on what you
    know about the movie </p>
</b>


<?php

mysql_connect('localhost', 'root', '');

mysql_select_db('database');
$sql="SELECT * FROM names";

$records=mysql_query($sql);
mysql_connect();

?>
<html>
<body>

<table width="600" border="1" cellpadding="1" cellspacing="1">
</table>

<tr>
<th>fname</th>
<th>sname</th>
<th>age</th>
<tr>

<?php

while($name=mysql_fetch_assoc($records))
{
    echo "<tr>";

    echo "<td>.$name.['fname'].</td>";
    echo "<td>.$name.['sname'].</td>";
    echo "<td>.$name.['age'].</td>";

    echo "</tr>";
}


?>

</BODY>
</HTML>

我正在尝试从数据库中复制信息并将其显示在我的网页上。任何帮助将不胜感激。

Answer 1

你正在做=iif(Fields!ACT.Value = "N/V", "No Color", iif(Fields!ACT.Value = "N/K", "No Color", iif(Fields!ACT.Value = "N/A", "No Color", iif(Fields!NUM.Value < Fields!SHORT.Value , "Yellow", iif(Fields!NUM.Value > Fields!INC.Value, "Orange", iif((isnothing(Fields!ACT.Value <> "N/A" and Fields!NUM.Value) or isnothing(Fields!ACT.Value <> "N/K" and Fields!NUM.value) or isnothing(Fields!ACT.Value <> "N/V" and Fields!NUM.Value)), "Purple", "No Color"))))))两次 - 第二次没有参数。因此，当您执行mysql_connect()时，您没有有效的资源。

从数据库获取信息时出现PHP错误

1 个答案: