Question

我遇到下面的代码有问题，它给了我一个语法错误，我在手册或网上找不到任何关于它的内容。关于如何让它运行起来你有什么想法？

第一次尝试：

<?php
require("../dbpass.php");    $types = array('Buyer','Seller','Buyer / Seller','Investor');
$typeslist = implode ("','", $types);

$sql = "SELECT * FROM contacts WHERE contacttype IN ('$typeslist') AND status = 'New' ORDER BY date DESC";
$result = mysqli_query($mysqli,$sql) or die ("Error: ".mysqli_error($mysqli));
while ($row = mysqli_fetch_array($result)) {

第二次尝试（在“IN”之后加上“=”）：

<?php
require("../dbpass.php");
$types = array('Buyer','Seller','Buyer / Seller','Investor');
$typeslist = implode ("','", $types);

$sql = "SELECT * FROM contacts WHERE contacttype = IN ('$typeslist') AND status = 'New' ORDER BY date DESC";
$result = mysqli_query($mysqli,$sql) or die ("Error: ".mysqli_error($mysqli));
while ($row = mysqli_fetch_array($result)) {

这是代码的其余部分：

$firstname = $row ['firstname'];     

echo'.$firstname.';
}
?>

错误：您的SQL语法出错;查看与您的MySQL服务器版本对应的手册，以便在'IN（'买方'，'卖方'，'买方/卖方'，'投资方'）附近使用正确的语法，并且状态='新'ORDER BY da'在线1

Answer 1

在你的状态之后你有一个;。

while ($row = mysqli_fetch_array($result)); {

应该是

while ($row = mysqli_fetch_array($result)) {

Answer 2

删除=。

contacttype IN

不

contacttype = IN

Answer 3

尝试此查询

$sql = "SELECT * FROM contacts WHERE contacttype IN('".$typeslist."') AND status = 'New' ORDER BY date DESC";

SELECT * FROM表WHERE行IN（'list'）

3 个答案: