Question

<?php

include 'db.php';

$subassetcategory = $_GET["subassetcategory"];

if ($subassetcategory == "all") {
    $sql = "SELECT * FROM asset_subasset";
}
else {
    $sql = "SELECT * FROM asset_subasset WHERE subassetcategory = '" . $subassetcategory . "'";
}

$result = mysql_query($sql) or die(mysql_error());

if ($row = mysql_fetch_array($result)) {

    echo "<table border='1'>
            <tr>
            <th>Asset ID</th>
            <th>Subasset Category</th>
            <th>Subasset Name</th>
            </tr>";

    while ($row = mysql_fetch_array($result)) {
        $subassetid = $row['subassetid'];
        $assetid = $row['assetid'];
        $subassetname = $row['subassetname'];
        $subassetcategory = $row['subassetcategory'];

        echo "<tr>";
        echo "<td>" . $assetid . "</td>";
        echo "<td>" . $subassetcategory . "</td>";
        echo "<td>" . $subassetname . "</td>";
        echo "<td>" . "<a href='viewassetdetail.php?subassetid=$subassetid&assetid=$assetid'>View<a/>" . "</td>";
        echo "</tr>";
    }
    echo "</table>";
}
else {
    echo "<br> No data found </br>";
}

mysql_close($connect);

?>

请帮帮我，我无法弄清楚为什么这段代码无法正常返回结果。它从数据库中可用的确切行返回少于1行（如果结果假设有2行，它只返回1）。

if ($row = mysql_fetch_array($result)) {

    echo "<table border='1'>
            <tr>
            <th>Asset ID</th>
            <th>Subasset Category</th>
            <th>Subasset Name</th>
            </tr>";

    while ($row = mysql_fetch_array($result)) {
        $subassetid = $row['subassetid'];
        $assetid = $row['assetid'];
        $subassetname = $row['subassetname'];
        $subassetcategory = $row['subassetcategory'];

        echo "<tr>";
        echo "<td>" . $assetid . "</td>";
        echo "<td>" . $subassetcategory . "</td>";
        echo "<td>" . $subassetname . "</td>";
        echo "<td>" . "<a href='viewassetdetail.php?subassetid=$subassetid&assetid=$assetid'>View<a/>" . "</td>";
        echo "</tr>";
    }
    echo "</table>";
}
else {
    echo "<br> No data found </br>";
}

mysql_close($connect);

?>

如果我刚刚写了这段代码，那就完美了。

$result = mysql_query($sql) or die(mysql_error());

echo "<table border='1'>
    <tr>
    <th>Asset ID</th>
    <th>Subasset Category</th>
    <th>Subasset Name</th>
    </tr>";

while ($row = mysql_fetch_array($result)) {
    $subassetid = $row['subassetid'];
    $assetid = $row['assetid'];
    $subassetname = $row['subassetname'];
    $subassetcategory = $row['subassetcategory'];

    echo "<tr>";
    echo "<td>" . $assetid . "</td>";
    echo "<td>" . $subassetcategory . "</td>";
    echo "<td>" . $subassetname . "</td>";
    echo "<td>" . "<a href='viewassetdetail.php?subassetid=$subassetid&assetid=$assetid'>View<a/>" . "</td>";
    echo "</tr>";
}
echo "</table>";

mysql_close($connect);

?>

但如果结果为空，我希望它显示“找不到数据”。那么，有什么帮助吗？我知道这很简单但我几天都找不到答案了。提前谢谢！

Answer 1

你基本上是在if语句中切掉第一个结果：

if ($row = mysql_fetch_array($result))

只需检查返回的行数以保持结果集的完整性：

if(mysql_num_rows($result) > 0)

Answer 2

您已使用mysql_fetch_array()检查是否有任何结果。你已经在那里消耗了一排。所以，这就是你缺少一行的地方。您可以使用mysql_num_rows()

检查结果中是否有任何值

Answer 3

If ($result->num_rows() > 0 ) { do your stuff } else { no results found }

Answer 4

更简化的方法是在你进入循环数组之前检查结果集。这样你可以跳过所有这些，如果它是空的，直接进入空对话。

Answer 5

而不是

if ($row = mysql_fetch_array($result))
{
  .....................
}

尝试使用它：

 $num_rows = mysql_num_rows($result); //returns the number of rows selected

 if($num_rows==0)
 {
      //no data
 }
 else
 {
    //your code
 }

结果从数据库中可用的确切行返回少于1（减1）

5 个答案: