php - SELECT * FROM table WHERE row IN ('list')

Question

我在使用下面的代码时遇到了问题，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。你对我如何让它运行有什么想法？

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”之后放置一个“=”）：

<?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 语法有错误；检查与您的 MySQL 服务器版本相对应的手册以获取正确的语法，以便在 'IN ('Buyer','Seller','Buyer/Seller','Investor') AND status = 'New' ORDER BY da' 行附近使用1

Answer 1

你有一段;时间后的情况。

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

应该

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

Answer 2

删除 =。

contacttype IN

不是

contacttype = IN

Answer 3

试试这个查询

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