我對下面的代碼有問題,它給我一個語法錯誤,我在手冊或網上找不到任何關於它的東西。關於如何讓它運行起來,你有什麼想法?SELECT * FROM table WHERE row IN('list')
第一嘗試:
<?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)) {
第二嘗試(把一個 「=」 後 「IN」):
<?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)) {
這是代碼的其餘部分:
$firstname = $row ['firstname'];
echo'.$firstname.';
}
?>
錯誤:您的SQL語法有錯誤;檢查與您的MySQL服務器版本相對應的手冊,在'IN('Buyer','Seller','Buyer/Seller','Investor')AND AND status ='New'ORDER BY da'at line附近使用正確的語法1
我們能有一個工作環節,看看這個錯誤嗎? – Charlie
你能告訴我們'echo $ sql'的輸出,或者至少是你得到的錯誤嗎? –
更新請參閱錯誤 – Josh