除了将新文本更新到数据库之外，我有一个页面似乎工作正常。每次我点击提交时，它都会返回到原始数据。请让我知道要成功查询旧数据并更新新数据还缺少什么。
if (isset($_POST["submit"]) && $_POST["submit"] == "Update Load")
{
for ($count = 1; $count <= 6; $count++)
{
$fields[$count] = "";
if (isset($_POST["field" . $count . ""]))
{
$fields[$count] = trim($_POST["field" . $count . ""]);
//echo $fields[$count] . "<br />";
}
}
$con = mysql_connect("", "", "");
mysql_select_db("", $con);
$carriername = mysql_real_escape_string($_POST['carriername']);
$contact = mysql_real_escape_string($_POST['contact']);
$phone = mysql_real_escape_string($_POST['phone']);
$rating = mysql_real_escape_string($_POST['rating']);
$info = mysql_real_escape_string($_POST['info']);
$insert = "UPDATE carrierinfo Set `carriername` = '$carriername', `contact` = '$contact', `phone` = '$phone', `rating` = '$rating', `info` = '$info' WHERE `id` = '$id';";
mysql_query($insert) or die(mysql_error());
$select = "SELECT `carriername` ,`contact` ,`phone` ,`rating` ,`info` ,`id` FROM `carrierinfo` ORDER BY `carriername` DESC;";
$result = mysql_query($select) or die(mysql_error());
}
if ($rating == "")
{
$rating = "3";
}
if (isset($_GET["id"]))
{
$con = mysql_connect("", "", "");
mysql_select_db("", $con);
$id = mysql_real_escape_string($_GET["id"]);
$select = "SELECT * FROM `carrierinfo` WHERE `id` = '$id'";
$result = mysql_query($select) or die(mysql_error());
$fields = mysql_fetch_array($result, MYSQL_BOTH);
mysql_close($con);
}
else
{
header("Location:board.php");
}
?>
</script>
<style ="text-align: center; margin-left: auto; margin-right: auto;"></style>
</head>
<body>
<div
style="border: 2px solid rgb(0, 0, 0); margin: 16px 20px 20px; width: 400px; background-color: rgb(236, 233, 216); text-align: center; float: left;">
<form action="" method="post";">
<div
style="margin: 8px auto auto; width: 300px; font-family: arial; text-align: left;"><br>
<table style="font-weight: normal; width: 100%; font-size: 12px;"
border="1" bordercolor="#929087" cellpadding="6" cellspacing="0">
<table
style="font-weight: normal; width: 100%; text-align: right; font-size: 12px;"
border="1" bordercolor="#929087" cellpadding="6" cellspacing="0">
<tbody>
<tr>
<td style="width: 35%;">Carrier Name:</td><td><input id="carriername" name="carriername" maxlength="50" style="width: 100%;" type="text" value="<?php echo $fields[carriername]; ?>">
</tr>
<tr>
<td style="width: 35%;">Contact:</td><td><input id="contact" name="contact" maxlength="50" style="width: 100%;" type="text" value="<?php echo $fields[contact]; ?>">
</tr>
<tr>
<td style="width: 35%;">Phone:</td><td><input id="phone" name="phone" maxlength="50" style="width: 100%;" type="text" value="<?php echo $fields[phone]; ?>">
</tr>
<tr>
<td style="width: 10%;">Carrier Rating:</td><td>
<select id="rating" name="rating">
<option value=""></option>
<option <?php if($fields[rating] == "1") echo "selected"; ?> value="1">1</option>
<option <?php if($fields[rating] == "2") echo "selected"; ?> value="2">2</option>
<option <?php if($fields[rating] == "3") echo "selected"; ?> value="3">3</option>
<option <?php if($fields[rating] == "4") echo "selected"; ?> value="4">4</option>
<option <?php if($fields[rating] == "5") echo "selected"; ?> value="5">5</option>
</select>
</td>
</tr>
<tr>
<td style="width: 10%;">Carrier Info:</td><td style="text-align: center;" colspan="2"><textarea name="info" maxlength="65535" style="width: 100%; height: 4em;"><?php echo $fields[info]; ?></textarea></td>
</tr>
</tbody>
</table>
<p style="text-align: center;"><input name="submit" value="Update"
class="submit" type="submit"></p>
</div>
</form>
<input type="button" onclick="window.location.href='test3.php';" value="Back" />
</div>
<p style="margin-bottom: -20px;"> </p>
</body>
最佳答案
$insert = "UPDATE carrierinfo Set `carriername` = '$carriername', `contact` = '$contact', `phone` = '$phone', `rating` = '$rating', `info` = '$info' **WHERE `id` = '$id';**";
在运行 UPDATE 查询之前，您尚未声明 $id 应该是什么，因此不会更新任何内容。
关于php - 在 sql 中插入和更新文本，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13316791/