我从mysql数据库中的两列中提取数据,并在线图中直观地表示它。下面的图表显示在我想放入textarea的网页上,该文本区域根据图表中显示的数据提供反馈信息。我想知道使用if else语句是否可行?我试图找出如果条件满足到textarea本身我将如何设置文本。它会是这样的吗?任何建议将不胜感激。
<?php
$MilkSolids = $row['milk_solids'];
$TagNumber= $row['tag_number'];
$average = '50';
$myquery = "SELECT `milk_solids`, `tag_number` FROM `milk` ";
$row = mysqli_fetch_array($myquery);
if($Milksolids > 50)
{
<textarea rows="2" cols="20">
<?php
echo "good feedback message"; ?>
</textarea>
}
else($Milksolids < $average)
{
<textarea rows="2" cols="20">
<?php
echo "bad feedback message"; ?>
</textarea>
}
?>
答案 0 :(得分:0)
我建议在PHP变量中构建标记,然后立即回显标记:
<?php
$average = '50';
$myquery = "SELECT `milk_solids`, `tag_number` FROM `milk` ";
$row = mysqli_fetch_array($myquery);
$MilkSolids = $row['milk_solids'];
$TagNumber= $row['tag_number'];
if ($row['milk_solids'] > 50) {
$msg = 'good feedback message';
}elseif ($Milksolids < $average) {
$msg = 'bad feedback message';
}else{
$msg = 'some other message';
}
$out = '<textarea rows="2" cols="20">' .$msg. '</textarea>';
echo $out;
?>
请注意,您必须在(1)mysql_query
和(2)mysql_fetch_assoc
语句之后放置这些行:
$MilkSolids = $row['milk_solids'];
$TagNumber= $row['tag_number'];