如何解决如何根据条件打开PHP表单If else方法
我试图根据两个变量'Type'和“ facility”的结果打开不同的表单。但是在我的代码中,无论何时单击查看按钮,它始终会打开Form1.php,而不管其条件如何。记录。我检查了我的代码,但找不到我的错误。我们将非常感谢您的帮助。
<?php
$type =$result->type;
$facility =$result->facility;
if ($facility =='C1' or $facility =='C2' or $facility =='C3' OR $facility ='C4' OR $facility =='C5'){
$formid='FORM1.php';
} elseif(($type =='T1' OR $type =='T2' OR $type =='T3') AND ($facility =='C6' or $facility =='C7' or $facility =='C8' OR $facility =='C9') ){
$formid='FORM2.php';
} elseif(($type =='T4' or $type =='T5' )AND ($facility =='C6' or $facility =='C7' or $facility =='C8' OR $facility =='C9') ){
$formid='FORM3.php';
}else{
$formid='nothing.php';
}
?>
<a onclick="window.open('<?php echo $formid?>?sampleid=<?php echo htmlentities($result->sampleid);?>','_blank','height=700,width=1200 top=200 left=350,location=no,toolbar=no,resizable=0,scrollbars=no')"><button class="btn btn-primary"><i class="fa fa-edit "></i> VIEW</button>
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