我一直在寻找一个答案,为什么这个工作几个小时都没有用,我很难过。
这是包含所涉及的javascript和表单的脚本。
<script language="Javascript" type="text/javascript">
function complete(init){
alert ("in function with " + init);
var aList = new Array(<?php echo $aList; ?>);
var iList = new Array(<?php echo $iList; ?>);
for (var i = 0; i < iList.length; i++){
if (init == iList[i]){
alert ("replacing " + init + " with " + aList[i]);
this.frmMain.txtAtty.value = aList[i];
}
}
}
</script>
<FORM METHOD="POST" NAME="frmMain" ACTION=<?php echo $_SERVER["PHP_SELF"]; ?>>
<table width="75%" align="center">
<tr>
<td width="25%" align="right">Name:</td>
<td>
<input type="text" name="txtSender" size="30" value=""/><span class="noteText"> Your Name</span>
</td>
</tr>
<tr>
<td width="25%" align="right">Attorney:</td>
<td>
<input type="text" name="txtAtty" size="30" value="" onblur = "complete(this.value)">
</td>
</tr>
两个PHP echo语句是Arrays的参数。完整的(this.value)函数应该采用3字母代码(在iList数组中)并用名称替换它。警报在那里用于调试目的,但是当我运行页面时我没有得到任何警报。有什么想法吗?
答案 0 :(得分:1)
this.frmMain 没有在任何地方定义
变化
onblur = "complete(this.value)"
到
onblur = "complete(this)"
并使用
function complete(field){
var init = field.value;
alert ("in function with " + init);
var aList = new Array(<?php echo $aList; ?>);
var iList = new Array(<?php echo $iList; ?>);
for (var i = 0; i < iList.length; i++){
if (init == iList[i]){
alert ("replacing " + init + " with " + aList[i]);
field.value = aList[i];
}
}
}