SELECT * FROM WHERE 查询未检索到任何结果

Question

我正在使用 SELECT * FROM "" WHERE "" = "" 查询，但我不确定自己做错了什么。我正在尝试 select 基于其 PO 的项目，它是 table 中一行的完全唯一标识符。这是我这样做的过程：

$jobnumber = $_GET['jref'];
$query = "SELECT * FROM `po_10152796` WHERE `po` = " .$jobnumber;
$results = mysqli_query($conn,$query) or die(mysqli_error($conn));
$rowitem = mysqli_fetch_array($results);
    $jobname = $rowitem['Job Name'];
    $phone = $rowitem['phone'];

我知道的是正确的：

"jobnumber" 被正确检索并与 table
table 名为 "po_10152796" 并且有一列名为 "po"

Answer 1

尝试使用这一行进行查询

$query = "SELECT * FROM `po_10152796` WHERE `po` = '" .$jobnumber. "' ";

Answer 2

忘记post这个，用准备好的语句重新编写代码并且它有效，不确定我到底改变了什么但无论如何它在这里：

    $jobnumber = $_GET['jref'];
$stmt = $conn->prepare( "SELECT `Job Name`, `Address`, `phone`, `description`, `materials` FROM po_10152796 WHERE po = ?");
$stmt->bind_param("i", $jobnumber);
if($stmt->execute()){
    $stmt->bind_result($jobname, $address, $phone, $description, $materials);
    $stmt->fetch();
}

SELECT * FROM WHERE 查询未检索到任何结果

SELECT * FROM WHERE Query isn't retrieving any results

mysql

select

mysqli

fetch