当我尝试运行此代码时,任何结果都没有显示。请帮助我
......有些代码在这里......
$result = mysql_query("select * from dataform where date between '" . $d1 . "' and '" . $d2 . "'");
if (!$result) {
echo 'No result';
}
else{
echo $d1;
echo $d2;
echo "<table class=\"hovertable\">";
echo "
<tr onmouseover=\"this.style.backgroundColor='#ffff66';\" onmouseout=\"this.style.backgroundColor='#d4e3e5';\">";
echo "<th>File No</th>";
echo "<th>Manufactor</th>";
echo "<th>Address</th>";
echo "<th>Supplier</th>";
echo "<th>Place Site</th>";
echo "<th>Tender Ref</th>";
echo "<th>Award No</th>";
echo "</tr>";
while ($row = mysql_fetch_array($result)) {
echo "<tr onmouseover=\"this.style.backgroundColor='#ffff66';\" onmouseout=\"this.style.backgroundColor='#d4e3e5';\">";
echo "<th>" . $row["fileno"] . "</th>";
echo "<th>" . $row["manufacture"] . "</th>";
echo "<th>" . $row["address"] . "</th>";
echo "<th>" . $row["sup"] . "</th>";
echo "<th>" . $row["placesite"] . "</th>";
echo "<th>" . $row["tenderref"] . "</th>";
echo "<th>" . $row["awardno"] . "</th>";
echo "</tr>";
}
答案 0 :(得分:4)
尝试使用反引号`转义列DATE
,因为它是MySQL数据类型。另一个建议是避免使用查询的字符串连接,因为它很容易注入mysql。使用 PHP PDO 或 MYSQLi 。
使用PDO的示例:
<?php
$stmt = $dbh->prepare("SELECT * FROM dataform WHERE `date` BETWEEN ? AND ?");
$stmt->bindParam(1, $d1);
$stmt->bindParam(2, $d2);
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC); //Fetch all results in form of associative array.
?>
请记住始终清理您的输入。
更新1
您没有在WHILE
循环中检查条件。你现在的做法是分配一个错误的值。尝试使用==
而不是
while ($row = mysql_fetch_array($result))
{
}
将其更改为此
while ($row == mysql_fetch_array($result))
{
}