doudouji2016 2013-03-16 18:07
浏览 20

如何将我的PHP变量传递给javascript

I have a dynamic dropdown. That populates the second drop down. My javascript is not taking my variables when I change my selection when I tested with one php variable. Am new to this

Javascript:

<script type="text/javascript">
function populate(s1,s2){
var s1 = document.getElementById(s1);
var s2 = document.getElementById(s2);
s2.innerHTML = "";
if(s1.value=="<?php echo $car; ?>"){
var optionArray=[" ","camero Camaro","convette Convette"];
} else
if(s1.value=="Dodge"){
var optionArray=[" ","avanger Avanger","challenger Challenger"];
} else if(s1.value=="Ford"){
var optionArray=[" ","mustang Mustang","fiesta Fiesta"];
}
for (var option in optionArray){
var pair = optionArray[option].split(" ");
var newOption = document.createElement("option");
newOption.value = pair[0];
newOption.innerHTML = pair[1];
s2.options.add(newOption);
}
}
</script>

PHP file:

<select name="slct1" id="slct1" onchange="populate(this.id,'slct2')">

<option value=""></option>
<option value="<?php $car = 'Chevy';echo $car; ?>">Chevy</option>
<option value="Dodge">Dodge</option>
<option value="Ford">Ford</option>
</select>
<hr />
<select id="slct2" name="slct2">
</select>
<hr />
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5条回答 默认 最新

  • drdyf42880 2013-03-16 18:10
    关注

    It is because you assign a variable <?php $car = 'Chevy'; ?>

    You need to print it <?php echo 'Chevy'; ?> or <?php echo $car; ?>

    评论

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