从php if/else语句的结果创建javascript操作
我有一段代码在PHP中运行良好:从php if/else语句的结果创建javascript操作,javascript,php,Javascript,Php,我有一段代码在PHP中运行良好: <?php $doUPDATE = mysqli_query($con, $sqlUPDATE); if (!$doUPDATE) { die('Could not update data: ' . mysqli_error($con)); } if ($OutcomeValue <> 'null') { $sqlMOVE = "INS
<?php
$doUPDATE = mysqli_query($con, $sqlUPDATE);
if (!$doUPDATE) {
die('Could not update data: ' . mysqli_error($con));
}
if ($OutcomeValue <> 'null') {
$sqlMOVE = "INSERT INTO results SELECT * FROM report WHERE LineID = $LineID ;" ;
$sqlDELETE = "DELETE FROM report where LineID = $LineID ;" ;
$doMOVE = mysqli_query($con, $sqlMOVE);
if (!$doMOVE)
{
die('Could not MOVE data: ' . mysqli_error($con));
}
$doDELETE = mysqli_query($con, $sqlDELETE);
if (!$doDELETE)
{
die('Could not DELETE data: ' . mysqli_error($con));
}
}
您不应该使用过程化PHP,现在是2014年,但这不是重点。您可以在php页面上执行脚本标记,如下所示:
<script type="text/javascript">
var blah = <?php echo $variable;?>
</script>
变量blah=
只需回显结果值,现在您可以在js中访问它,您可以执行以下操作:
if (!$doMove) {
<script type="text/javascript">
alert(<?php echo json_encode(mysqli_error($con)); ?>);
</script>
}
if(!$doMove){
警惕();
}