Php 类mysqli_result的对象无法转换为字符串5
目标Php 类mysqli_result的对象无法转换为字符串5,php,mysqli,Php,Mysqli,目标 用户在文本字段中输入一个数字并点击计算 然后用250减去这个数字 根据$result是什么,将显示特定的图像 问题 当页面运行时,我收到以下错误消息 可捕获的致命错误:在第95行的/home/cs12jcw/public_html/n-power/includes/calculator.php中,类mysqli_result的对象无法转换为字符串 第95行是 echo "<img src='$image' alt='' />"; echo”“; 完
echo "<img src='$image' alt='' />";
echo”“;
完整代码
<?php
$valuea = (isset($_POST['valuea']) && is_numeric($_POST['valuea'])) ? $_POST['valuea'] : 0;
$valueb = 250;
$answer = $valuea - $valueb;
?>
<form method='post' action='calculator.php'>
<table border='0' width='500px' cellpadding='3' cellspacing='1' class="table">
<tr class="calcheading">
<td colspan="2"><strong>Work out how much you could be saving</strong></td>
</tr>
<tr class="calcrow">
<td>How much do you spend a year?</td>
<td align="center"><input type='text' name='valuea' value="$valuea"/></td>
</tr>
<tr class="calcrow">
<td>Minus the average price of an n-power student tarrif* Leave Blank:</td>
<td align="center"><input type='text' name='valueb' value="$valueb"/></td>
</tr>
<tr class="submit">
<td colspan="2"><input type='submit' value='Calculate'/></td>
</tr>
<tr class="calcrow">
<td><i>You could be saving:</td>
<td align="center"><input type="text" value="<?php echo round($answer)?>"></td></i>
</tr>
</table>
</form>
<?php
if($db_server){
switch( $answer ){
case $answer > 0 and $answer < 150 : $image = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = 'image1'");
break;
case $answer < 250 : $image = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = 'image2'");
break;
case $answer < 350 : $image = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = 'image3'");
break;
case $answer < 450 : $image = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = 'image4'");
break;
case $answer < 550 : $image = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = 'image5'");
break;
}
echo "<img src='$image' alt='' />";
}
?>
计算出你可以节省多少钱
你一年花多少钱?
减去n-power学生tarrif*的平均价格留空:
你可以节省:
您忘记获取结果:
$row = mysqli_fetch_array($image);
echo "<img src='$row[URL]' alt='' />";
$row=mysqli\u fetch\u数组($image);
回声“;
您正在将$image变量设置为mysql结果对象。您仍然需要从结果中获取数据。试试下面的方法
$result = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = 'image1'");
$obj = mysqli_fetch_object($result);
$image = $obj->URL;
另外,您可以将switch语句修改为下面的语句,这样您只有一个点发出查询
$imagename = '';
switch( $answer ){
case $answer > 0 and $answer < 150 : $imagename = 'image1';
break;
case $answer < 250 : $imagename = 'image2';
break;
case $answer < 350 : $imagename = 'image3';
break;
case $answer < 450 : $imagename = 'image4';
break;
case $answer < 550 : $imagename = 'image5';
break;
}
if ($imagename) {
$result = mysqli_query($db_server, "SELECT URL FROM images WHERE imagename = '$imagename'");
$obj = mysqli_fetch_object($result);
$image = $obj->URL;
echo "<img src='$image' alt='' />";
}
$imagename='';
开关($应答){
案例$answer>0和$answer<150:$imagename='image1';
打破
案例$answer<250:$imagename='image2';
打破
案例$answer<350:$imagename='image3';
打破
案例$answer<450:$imagename='image4';
打破
案例$answer<550:$imagename='image5';
打破
}
如果($imagename){
$result=mysqli_查询($db_服务器,“从图像中选择URL,其中imagename='$imagename');
$obj=mysqli\u fetch\u对象($result);
$image=$obj->URL;
回声“;
}
您显然需要重新学习PHP。检查输入[name=valueB]
的值属性的声明,您需要
围绕它。我试过了,但后来我得到一个错误,它说出于某种原因出现了意外?>此讨论完全针对一个单独的问题。