PHP/SQL脚本无法正常工作

Question

我正在尝试生成前10个标签,我认为这非常简单.

    //generate 10 top tags
$tagSQL = mysql_fetch_array(mysql_query("SELECT * FROM tags"));
$topArray = array();
foreach($tagSQL as $poland)
{
    if($poland["tagID"] == 1)
    {
        $topArray[0] = $topArray[1] + 1;
    }
    if($poland["tagID"] == 2)
    {
        $topArray[1] = $topArray[2] + 1;
    }
    if($poland["tagID"] == 3)
    {
        $topArray[2] = $topArray[3] + 1;
    }
    if($poland["tagID"] == 4)
    {
        $topArray[3] = $topArray[4] + 1;
    }
}
function printTopTags()
{
    $n = 0;
    foreach($topArray as $buddha)
    {
        $n = $n + 1;
        if(sizeOf($topArray) > $n)
        {
            $hersheyBar = " ";
        }
        else
        {
            $hersheyBar = "";
        }
        $finalFinalEndArray = mysql_fetch_array(mysql_query("SELECT tagName FROM tags WHERE tagID = '$buddha'");
        foreach($finalFinalEndArray as $waterBottle)
        {
            echo $waterBottle . $hersheyBar;
        }
    }   
}

我遇到的唯一问题是它认为我在代码中的某处有语法错误,但无论哪行都省略了语法错误.

另外,是否有一种更简单的方法来处理10个不同阵列点的if语句,而不是四个？

Answer 1

您在本声明的末尾缺少")"

它应该是

$finalFinalEndArray = mysql_fetch_array(mysql_query("SELECT tagName FROM tags WHERE tagID = '$buddha'"));