JQuery的/阿贾克斯接受真/假的PHP函数 [英] JQuery/Ajax accepting True/False from PHP function

问题描述

function check_jobref_availability(){
    if ($job_reference_already_exists == 0) { 
        $output = TRUE;
    } else {
        $output = FALSE;
    }
    echo $output;
}

功能齐全

function check_jobref_availability(){       
    $output = 1;       
    $jobrefference = $this->uri->segment(3); //mysql_real_escape_string($_REQUEST['ptitle']);
    $strSQL = "SELECT count(*) FROM projects where ptitle = '" . $jobrefference . "' and companyid = 1";
    $job_reference_already_exists = get_singlecolumn($strSQL);        

    //if it is zero means. no refrence was found and it is good to go to add the new job reference.
    if ($job_reference_already_exists == 0) { 
        $output = 1; // NOT FOUND SO GO AHED 
    } else {
        $output = 0; // MEANS ALREAYD EXISTS give an alert
    }
    echo $output;

}

这是我的PHP code（在code实际查询数据库的其余部分），并在此基础上，我需要返回true或false返回0或1。

This is my PHP Code (rest of the code actually queries the DB) and returns 0 or 1. Based on which i need to return true or false.

var newjobrefrence = $('#newjobrefrence').val().trim();
$.ajax(
{
    type: "POST",
    url: "<?php echo base_url(); ?>postproject/check_jobref_availability/" + newjobrefrence,        
    data: {ptitle: newjobrefrence},
    success: function(response){
        if (response.d == "false") {
            jobrefrencealreadyexists = 1;
            alert("Job ref found");

        }else{
           jobrefrencealreadyexists = 0;
           alert("Not found");
        }

    },
    error: function(xhr, ajaxOptions, thrownError) {
        alert(xhr.status);
        alert(thrownError);
    }
});

这是我的jQuery / AJAX调用。我是新来的jQuery / AJAX。

And this is my jquery/ajax call. I am new to jquery/ajax.

所有它需要做的就是调用check_jobref_availability，并根据返回值做一个警告。

All it needs to do is call check_jobref_availability, and based on the return value do an alert.

我已经试过好像回到杰森（我不是100％肯定，如果正确地做，要么），或返回真或假和简单的真不同组合;和虚假的;但我不能看到我的jQuery ..

I have tried different combinations like returning Jason (I am not 100% sure if was doing that correctly either), or returning "True" or "False" and simple true; and false; but I can not see the response in my jquery..

我有离开我的头，我一直在整个周末拉扯他们在这几毛。

I have got few hair left on my head as I have been pulling them all weekend on this.

任何帮助将是pciated救我的发型多少AP $ P $。

Any help would be much appreciated to save my hair style.

推荐答案

试试这个一次，并检查您的 Firebug的净面板 AJAX调用的响应：

Try this once, and check your firebug's net panel for the ajax call's response:

function check_jobref_availability(){
if ($job_reference_already_exists == 0) { 
    $output = 1;
} else {
    $output = 0;
}
echo $output;
}

success: function(response){
    if (response == 1) {
        jobrefrencealreadyexists = 1;
        alert("Job ref found");
    }else{
       jobrefrencealreadyexists = 0;
       alert("Not found");
    }

}

如果它不列入帮助，请点击浏览器此链接直接看到你的控制器的输出如果回声（ING）正确的输出：

If it doesnot helps, try hitting this link in your browser directly to see the output of your controller if its echo(ing) correct output:

base_url()/postproject/check_jobref_availability/[any_job_refrence_here]

更新

var newjobrefrence = $('#newjobrefrence').val().trim();
$.ajax({
    type        : "POST",
    url         : "<?php echo base_url(); ?>postproject/check_jobref_availability/", //removed the uri segment       
    data        : { newjobrefrence : newjobrefrence },
    success     : function(response){
        if (response == "1") {
            jobrefrencealreadyexists = 1;
            alert("Job ref found");

        }else{
           jobrefrencealreadyexists = 0;
           alert("Not found");
        }

    },
    error       : function(xhr, ajaxOptions, thrownError) {
        alert(xhr.status);
        alert(thrownError);
    }
});

function check_jobref_availability(){
    if( $this->input->post(null) ){
        $id = $this->input->post('newjobrefrence'); #fetch from the post array and not the uri segment
        //your query here with $job_reference_already_exists
        /*if ($job_reference_already_exists == 0) { 
            $output = 1;
        } else {
            $output = 0;
        }*/
        echo $job_reference_already_exists;
    }

}

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