SQL 语法错误答案

【问题标题】：Error in SQL syntaxSQL 语法错误
【发布时间】：2015-03-22 02:51:42
【问题描述】：

$query = "SELECT * FROM guest WHERE ";
if ($_POST['include2'] == "yes") {
$guestNameLast = isset($_POST['guestNameLast']) ? $_POST['guestNameLast'] :     '';
$query .= "guestNameLast LIKE '%$guestNameLast%' ";
$msg .="guest that <b>" . $guestNameLast . "</b><br/>";

您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，以在第 1 行的 'guestNameLast LIKE '%%'' 附近使用正确的语法。如何解决此错误？

【问题讨论】：

在 $query 运行之前打印/记录它，这样您就可以看到生成的 SQL 是什么

标签： php mysql sql syntax sql-like

【解决方案1】：

在 If 条件中添加 WHERE 子句和 AND 还检查 $_POST['guestNameLast'] 不为空

$query = "SELECT * FROM guest WHERE 1 ";
if ($_POST['include2'] == "yes") 
{
if(isset($_POST['guestNameLast']) && ($_POST['guestNameLast'] != ""))
 {
 $guestNameLast = trim($_POST['guestNameLast']);
$query .= " AND guestNameLast LIKE '%$guestNameLast%' ";
 }


$msg .="guest that <b>" . $guestNameLast . "</b><br/>";

【讨论】：

【解决方案2】：

如果只有guestNameLast，则添加 WHERE 子句

$query = "SELECT * FROM guest";

if ($_POST['include2'] == "yes") {

$guestNameLast = '';
if(isset($_POST['guestNameLast']) && trim($_POST['guestNameLast']))
{
    $guestNameLast = trim($_POST['guestNameLast']);
    $query .= " WHERE  guestNameLast LIKE '%$guestNameLast%' ";
}

$msg .="guest that <b>" . $guestNameLast . "</b><br/>";

【讨论】：

【解决方案3】：

在 where 子句中添加条件。所以如果$_POST['guestNameLast'] 为空，它不会抛出错误：

$query = "SELECT * FROM guest WHERE 1=1 ";
if ($_POST['include2'] == "yes") {
$guestNameLast = isset($_POST['guestNameLast']) ? $_POST['guestNameLast'] :     '';
$query .= " AND guestNameLast LIKE '%$guestNameLast%' ";
$msg .="guest that <b>" . $guestNameLast . "</b><br/>";

【讨论】：

【解决方案4】：

错误即将到来，因为如果post 数据不可用，则WHERE 将出现在查询中。尝试使用 -

$query = "SELECT * FROM guest ";
if ($_POST['include2'] == "yes") {
$guestNameLast = isset($_POST['guestNameLast']) ? $_POST['guestNameLast'] :     '';
$query .= " WHERE ";
$query .= "guestNameLast LIKE '%$guestNameLast%' ";
$msg .="guest that <b>" . $guestNameLast . "</b><br/>";

【讨论】：