Question

以下SQL语句的update语句无效。

执行后说：

Query performed correctly: 0 row(s) affected.

每次调用后，timescount列应增加1，但不会发生。我需要你专业的眼睛仔细审查这个陈述。我还在代码下面显示了我的数据库模式。

require '15_12.php';

//Now, let us create an instance of mydb.
$mydb = new mydb ("localhost","root","");

//Select a database to use.
$mydb->selectdb ("myquotes");

//Then, let's try to return a row set.
$adata = $mydb->getrows ("SELECT url_string FROM tshirt WHERE timescount = 0 ORDER BY url_id ASC LIMIT 3");

    for ($i = 0; $i < count ($adata); $i++){

        $result = PIPHP_GetLinksFromURL($adata[$i]);
        echo "<ul>";
        for ($j = 0 ; $j < count($result) ; ++$j) {
        echo "<li>$result[$j]</li>";
        }

        //Now, hold the row in a variable
        $url_escaped = $adata[$i];

        //Now, let's perform an action.
        $mydb->execute ("UPDATE tshirt SET timescount=timescount+1 WHERE url_string='url_escaped'");



    }

数据库结构

CREATE TABLE IF NOT EXISTS `tshirt` (
  `url_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `url_string` varchar(2048) NOT NULL,
  `timescount` int(11) unsigned NOT NULL,
  `dt` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`url_id`)
)

Answer 1

你错过了变量中的$，导致它被用作字符串文字：

$mydb->execute ("UPDATE tshirt SET timescount=timescount+1 WHERE url_string='url_escaped'");
                                                                           ^^^^
                                                                           HERE

应该是

$mydb->execute ("UPDATE tshirt SET timescount=timescount+1 WHERE url_string='$url_escaped'");

SQL查询的一部分未执行

1 个答案: