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Jsf 2 Primefaces微调器最大属性达到重置值_Jsf 2_Primefaces - Fatal编程技术网

Jsf 2 Primefaces微调器最大属性达到重置值

Jsf 2 Primefaces微调器最大属性达到重置值,jsf-2,primefaces,Jsf 2,Primefaces,在微调器中,我使用最小值和最大值属性。在最小值=0和最大值=59中。当我单击微调器时,它将增加值,以表明我对其再次达到最大值表示怀疑。单击“不增加”时,它将再次启动最小值。有可能吗?使用ajax是一种方法: 在您的页面中: <h:form id="form"> <p:spinner min="0" max="60" value="#{yourBean.cuenta}" onkeypress="return false"> <p:ajax

在微调器中,我使用最小值和最大值属性。在最小值=0和最大值=59中。当我单击微调器时,它将增加值,以表明我对其再次达到最大值表示怀疑。单击“不增加”时,它将再次启动最小值。有可能吗?

使用ajax是一种方法:

在您的页面中:

<h:form id="form">

    <p:spinner min="0" max="60" value="#{yourBean.cuenta}"  onkeypress="return false">
        <p:ajax process="@this" update="@this" listener="#{yourBean.cuentaF}"/>
    </p:spinner>

</h:form>
<h:body>
<!-- THE SCRIPT -->
<script>
  $(document).ready(function() {

        var previous = -1;
        var $htmlvalue = $('#form\\:jss .ui-spinner-input');
        var $uparrow = $('#form\\:jss .ui-spinner-up');
        var $downarrow = $('#form\\:jss .ui-spinner-down');
        $htmlvalue.val(0);
        PF('vjss').value=0;

        $uparrow.click(function(){

            var limit = PF('vjss').value == previous;               

            if(PF('vjss').value==59 &amp;&amp; limit){
                PF('vjss').value=0; 
                $htmlvalue.val(0);
            }
            previous = PF('vjss').value;
            });

        $downarrow.click(function(){

            var limit = PF('vjss').value == previous;

            if(PF('vjss').value==0 &amp;&amp; limit){
                PF('vjss').value=59; 
                $htmlvalue.val(59);
            }               
            previous = PF('vjss').value;
            });         
  });       
</script>

<!-- THE JSF SPINNER COMPONENT -->
<h:form id="form">
    <p:spinner widgetVar="vjss" id="jss" min="0" max="59" value="#{yourBean.cuenta}"/>
</h:form>
</h:body>
<p:spinner id="ajaxSpinner" min="0" max="59" value="#{yourBean.number}">
    <p:ajax update="@this" process="@this" />
</p:spinner>
编辑:

使用jQuery是另一种方式:

在您的页面中:

<h:form id="form">

    <p:spinner min="0" max="60" value="#{yourBean.cuenta}"  onkeypress="return false">
        <p:ajax process="@this" update="@this" listener="#{yourBean.cuentaF}"/>
    </p:spinner>

</h:form>
<h:body>
<!-- THE SCRIPT -->
<script>
  $(document).ready(function() {

        var previous = -1;
        var $htmlvalue = $('#form\\:jss .ui-spinner-input');
        var $uparrow = $('#form\\:jss .ui-spinner-up');
        var $downarrow = $('#form\\:jss .ui-spinner-down');
        $htmlvalue.val(0);
        PF('vjss').value=0;

        $uparrow.click(function(){

            var limit = PF('vjss').value == previous;               

            if(PF('vjss').value==59 &amp;&amp; limit){
                PF('vjss').value=0; 
                $htmlvalue.val(0);
            }
            previous = PF('vjss').value;
            });

        $downarrow.click(function(){

            var limit = PF('vjss').value == previous;

            if(PF('vjss').value==0 &amp;&amp; limit){
                PF('vjss').value=59; 
                $htmlvalue.val(59);
            }               
            previous = PF('vjss').value;
            });         
  });       
</script>

<!-- THE JSF SPINNER COMPONENT -->
<h:form id="form">
    <p:spinner widgetVar="vjss" id="jss" min="0" max="59" value="#{yourBean.cuenta}"/>
</h:form>
</h:body>
<p:spinner id="ajaxSpinner" min="0" max="59" value="#{yourBean.number}">
    <p:ajax update="@this" process="@this" />
</p:spinner>

与@lametaweb类似的解决方案,但不改变最大值和最小值。使用这两种解决方案,您几乎不需要向BackBean添加任何代码

在您的页面中:

<h:form id="form">

    <p:spinner min="0" max="60" value="#{yourBean.cuenta}"  onkeypress="return false">
        <p:ajax process="@this" update="@this" listener="#{yourBean.cuentaF}"/>
    </p:spinner>

</h:form>
<h:body>
<!-- THE SCRIPT -->
<script>
  $(document).ready(function() {

        var previous = -1;
        var $htmlvalue = $('#form\\:jss .ui-spinner-input');
        var $uparrow = $('#form\\:jss .ui-spinner-up');
        var $downarrow = $('#form\\:jss .ui-spinner-down');
        $htmlvalue.val(0);
        PF('vjss').value=0;

        $uparrow.click(function(){

            var limit = PF('vjss').value == previous;               

            if(PF('vjss').value==59 &amp;&amp; limit){
                PF('vjss').value=0; 
                $htmlvalue.val(0);
            }
            previous = PF('vjss').value;
            });

        $downarrow.click(function(){

            var limit = PF('vjss').value == previous;

            if(PF('vjss').value==0 &amp;&amp; limit){
                PF('vjss').value=59; 
                $htmlvalue.val(59);
            }               
            previous = PF('vjss').value;
            });         
  });       
</script>

<!-- THE JSF SPINNER COMPONENT -->
<h:form id="form">
    <p:spinner widgetVar="vjss" id="jss" min="0" max="59" value="#{yourBean.cuenta}"/>
</h:form>
</h:body>
<p:spinner id="ajaxSpinner" min="0" max="59" value="#{yourBean.number}">
    <p:ajax update="@this" process="@this" />
</p:spinner>

在你的背景中:

private static final int MAX=59;
private static final int MIN=0;
private int number;

public int getNumber() {
    return number;
}

public void setNumber(int number) {

    if (this.number == MAX && number + 1 > MAX){
        this.number = MIN;
    }else if (this.number == MIN && number -1 < MIN){
        this.number=MAX;
    }else{
        this.number=number;
    }
}
private static final int MAX=59;
私有静态最终int MIN=0;
私有整数;
public int getNumber(){
返回号码;
}
公共无效集合号(整数){
如果(this.number==MAX&&number+1>MAX){
这个数字=MIN;
}else if(this.number==MIN&&number-1
可以处理javaScript,因为我允许键事件并尝试减少backbean事件,我编辑了初始解决方案。你试过第二种解决方案吗?它是一个用javascript实现的客户端解决方案。